ANALYSIS OF FUNCTIONS (D COURSE - PART II MATHEMATICAL TRIPOS)

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1 ANALYSIS OF FUNCTIONS (D COURS - PART II MATHMATICAL TRIPOS) CLÉMNT MOUHOT - C.MOUHOT@DPMMS.CAM.AC.UK Contents. Integration of functions 2. Vector spaces of functions 9 3. Fourier decomposition of functions Generalised derivatives of functions 27 References 33 Preamble. These notes are inspired by the boos of Lieb & Loss [4], Brézis [], Kolmogorov-Fomin [2, 3] and Rudin [5]. There is no claim of originality in the material presented here which is standard, apart from the arrangement and presentation. There will be three example sheets and a moc exam. xercises are included along the notes to help with understanding the material, their answers are not part of the examinable lecture notes.. Integration of functions Structure of the Tripos imposes constraints. We tae for granted the construction of the Lebesgue measure as well as familiarity with abstract measure theory. This is covered in the course Probability & Measure or the boos already mentioned. We focus on the applications to the analysis of functions... Recall: Lebesgue measure and integration. Measure theory invention (Borel, Lebesgue, Radon, Fréchet... ) was motivated by defining length and volume for more and more complicated sets; Lebesgue integration theory relies on measure theory and its invention was motivated by solving some theoretical and practical defiencies in Riemann integration theory. xercise. Recall that a function is Riemann-integrable if lower and upper Darboux sums both converge, to the same limit. Show that the pointwise limit of a sequence of Riemann-integrable functions is not necessarily Riemann-integrable.... Recalls on measure theory. Consider a set, and P() denotes the set of its subsets. Definition. (Algebra). A P() is an algebra is if it (i) is stable by finite union, (ii) is stable by absolute difference (A A \ A A), (iii) contains the whole set. Definition.2 (σ-algebra). A P() is a σ-algebra is if it (i) is stable by countable union, (ii) is stable by absolute difference, (iii) contains the whole set. (, A) is then a measurable space. Remar.3. () Observe that, by difference, algebra are stable by finite intersection and σ-algebra are stable by countable intersection. (2) Compare with the definition of a topology T P() on : contains and, stable by any union, finite intersections. The property of begin a σ-algebra is stable by intersection, therefore there is a notion of smallest σ- algebra containing a given collection of subsets M P(). When M = T is the topology on, the resulting σ-algebra is the Borel sets, denoted B(). B(R n ) is the smaller σ-algebra containing open balls. Date: Lent 208.

2 2 C. MOUHOT Definition.4 (Measure). A measure on (, A) is an application µ : A [0, + ] s.t. µ( ) = 0 and µ is σ-additive, i.e. countably additive: for (A ) in A, then µ( A ) = µ(a ). (, A, µ) is then a measure space. It is said complete if A A with µ(a) = 0 and B A implies B A (and µ(b) = 0). The completion of the Borel sets B(R n ) are the Lebesgue sets, denoted L(R n ). xercise 2. Construct a finitely additive µ with µ( ) = 0 that is not σ-additive () on subsets (to be defined) of N, (2) on the Borel sets of an interval of R. Prove that the finite additivity and the continuity from below (µ( n = A ) µ( = A n)) imply σ-additivity. Prove that the finite additivity, µ() < + and the continuity from above (µ( n = A ) µ( = A )) imply σ-additivity. xercise 3. Prove that the cardinal of B(R n ) is the same as R, and the cardinal of L(R n ) is the same as P(R) (hint: construct a zero-measure Cantor set and show any subset is a Lebesgue set). Theorem (xistence of the Lebesgue mesure) There exists a unique measure on (R n, B(R n )) s.t. µ( n i= [a i, b i ]) = n i= (b i a i ) (measure of hypercubes). Remar.5. This measure is σ-finite: there is a countable increasing covering sequence of sets with finite measures (e.g. [ N, N] n, N N ). This is useful in proofs for reducing to a finite measure set. Definition.6 (Measurable functions). Consider two measurable spaces (, A), (Y, B), then a function f : Y is measurable if for any B B then f (B) A. xercise 4. When B is the Borel sets, prove that it is enough to chec the definition for any B T open set. We then call the resulting measurable function a f Borel function. Prove also that composition, coordinate concatenation, coordinate restriction preserves measurability. Proposition.7 (Sequence of measurable functions). Consider two measurable spaces (, A), (Y, B) with Y metric space and B the Borel sets of Y. Consider a sequence of measurable functions f : Y s.t. f converges pointwise to f. Then f is measurable. Proof. Since B can be formed from open sets through the operations of countable union, countable intersection, and difference, it is enough to chec that f (U) A for any U T Y. Define U n := {x Y : d Y (x, Y \ U) > /n}, F n := {x Y : d Y (x, Y \ U) /n}. Observe that U n is open, F n is closed, and U = n U n = n F n with U n F n U n+ F n+ U. Then using the preimage of union is the union of preimages: f (U) = f U n = f (U n ) n n n l l f (U n) where the last set inclusion follows from the fact that U n is open. Then we have l l f (U n) l l f (F n) and by closure of F n we have l l f (F n) f (F n ), therefore f (U) = f (U n ) f (U n) f (F n) f (F n ) = f (U) n n n n l l l l which implies set equality and finally (countable union and intersection) that f (U) A. xercise 5. () When (Y, B) = (R, B(R)), adapt the proof and generalise the limit to limsup and liminf. (2) Is the statement still true when Y is a general topological space with a σ-algebra?

3 ANALYSIS OF FUNCTIONS 3.2. Integrability and convergence theorems. Riemann integration theory is based on subdivising the input space thans to the total order of the real line (whereas the output space can be a general Banach space). By contrast Lebesgue integration theory is based on subdivising the output space with the order of the the real line whereas the input space can be a general measure space. The respective approximation tools to go from discrete to continuous analysis are () Riemann/Darboux sums on the one hand and (2) simple functions on the other hand. Definition.8 (Simple functions). A function f : (, A) (R, B(R)) is simple if it is measurable and taes a finite number of values in [0, + ). Remar.9. The characteristic function of a set A is denoted χ A (returns one on A, zero elsewhere). Chec that χ A : (, A) (R, B(R)) is measurable iff A A. If so, it is a simple function. Proposition.0. Consider f : (, A) [0, + ] measurable where [0, + ] = [0, + ) {+ } is metrised with d(x, y) = arctan x arctan y (its topology includes the (a, + ] around + ) and endowed with the corresponding Borel sets. There is s non-decreasing sequence of simple functions converging pointwise to f. Proof. Define for any n : B n := {x f(x) n} and A i n := {x f(x) [(i )/2 n, i/2 n )} for i =,..., n2 n. These sets are all in A and partition. Define s n so that s n (i )/2 n on A i n and s n n on B n. To prove monotonicity observe that A i n = A 2i n+ A 2i+ n+. To prove convergence observe that on {f n}, the construction imposes f s n 2 n. Definition. (Integral of simple functions). Consider a measured space (, A, µ). The integral of a simple function s := n i= α iχ Ai, α i [0, + ), A i A on the set A is defined as n s dµ := α i µ(a i ). Remar.2. Observe then that the map A s dµ is a measure on (, A). i= Definition.3 (Integral of positive measurable functions). Let f : (, A, µ) [0, + ] measurable where [0, + ] is endowed with the Borel sets described above. Its integral on A is defined as { } f dµ := sup s dµ s simple function and s f [0, + ]. Remar.4. Observe that this integral of measurable functions valued in [0, + ] always maes sense in [0, + ]. Observe also that if has zero measure then f dµ = 0 ( in spite of possibly infinite values). xercise 6. () Prove the linearity of the integral from this definition (αf +βg) dµ = α f dµ+β g dµ. (2) Prove Chebyshev s inequality from this definition: µ({x f(x) α}) α f dµ. (3) Prove that if f : [0, + ] is measurable with finite integral then µ({x f(x) = + }) = 0. Theorem.5 (Beppo Levi - Lebesgue s monotone convergence theorem). Consider an increasing sequence of measurable functions f : (, A, µ) [0, + ] (f f + ) that converges pointwise to f, then A, f dµ = f dµ. lim + Proof. By considering f χ it is enough to consider =. The sequence f dµ of [0, ] is increasing therefore converges to some α [0, + ]. Since f dµ f dµ by monotonicity, this limit α f dµ. Let s a simple function below f and c (0, ). Define := {x f (x) cs(x)}. Chec that A measurable, +, and =. By continuity from below of the measure A A s dµ, we A have s dµ = lim s dµ. Tae + and c in f dµ f dµ cs dµ = c s dµ. to get α s dµ. Taing the supremum over s f gives α f dµ, which concludes the proof. xercise 7. () For any sequence of measurable functions f : (, A, µ) [0, + ] prove that ( f ) dµ = f dµ. (2) Prove that for f : (, A, µ) [0, + ] measurable, ν : A A f dµ is a measure, A and that for any g : (, A, µ) [0, + ] measurable, g dν = fg dµ.

4 4 C. MOUHOT Theorem.6 ( Fatou s lemma ). Given any sequence of measurable functions f : (, A, µ) [0, + ] ( ) (lim inf f ) dµ lim inf f dµ. Proof. The lim inf f is obtained as the pointwise limit of the non-decreasing sequence of functions F = inf{f l : l }. The latter are valued in [0, + ] and measurable using for instance {F a} = l {f l a}. The Beppo-Levi theorem applies: ( ) (lim inf f ) dµ = (lim F ) dµ = lim F = lim inf f l dµ. l Then use inf l f l dµ inf l f l dµ to conclude ( (lim inf f ) dµ lim inf l ) f l dµ = lim inf f dµ. Definition.7 (Integral of real or complex valued functions). A measurable function f : (, A, µ) C is integrable if f : [0, + ) (whose integral is defined above) satisfies f dµ < +. Its integral is then computed by splitting real/imaginary and positive/negative parts. Theorem.8 (Lebesgue s dominated convergence theorem). Let f : (, A, µ) C be a sequence of measurable functions that converges pointwise to f (convergence). We assume (domination) that there is g : (, A, µ) [0, + ] measurable and with finite integral so that f g for all. Then f and f are integrable and f f dµ 0. Proof. The domination implies the integrability of f and f. Consider h := 2g f f 0 valued in [0, + ] which converges pointwise to 2g. Use Fatou s lemma: 2g dµ = lim inf h dµ lim inf h = 2g dµ lim sup f f dµ which implies lim sup f f dµ = 0 and concludes the proof. Remar.9. The previous results extend when the pointwise convergence is replaced by the pointwise convergence almost everywhere on some A with µ( \ ) = 0: replace f by fχ. xercise 8. Formulate and prove continuity/differentiability of t F (t, ) dµ when F satisfies proper continuity/differentiability and domination assumption. Remar.20. In summary the construction of the theory of integration of complex-valued function has followed the standard process : () characteristic functions using the base measure, (2) simple functions by linear combination, (3) positive functions by Beppo-Levi, (4) real or complex-valued functions through the modulus. This standard process is used in many proofs. xercise 9. () Prove that for a bounded function on [a, b], Riemann-integrability implies Lebesgue-measurability (for the Lesbesgue σ-algebra) but not necessarily Borel-measurability, and implies Lebesgue-integrability with both integrals agreeing. (2) Prove that the Riemann-integrability is equivalent to the set of discontinuity having zero Lebesgue measure. [Cf. xercise 7 of xample sheet.].3. Lebesgue spaces: completeness, separability. Definition.2. We denote, for p [, + ], L p () the set of equivalence classes of measurable functions f : R (or C) such that f p is integrable (resp. f essentially bounded when p = +, i.e. bounded outside a null set), for the relation of almost everywhere equality. Theorem.22 (Riesz-Fischer). ndowed with f L p () := ( f p dµ) /p (resp. the essential supremum f L () := inf{m 0 s.t. µ({x f(x) M}) = 0} when p = + ), the space L p () is a Banach space (i.e. a complete normed vector space). xercise 0. Suppose f L () is supported on a set of finite measure, then prove that f L p () for all p [, + ), and f L p f L as p +.

5 ANALYSIS OF FUNCTIONS 5 Proof. Standard properties (in particular Minowsi inequality) are left to the reader. We prove the completeness. Assume first p [, + ): We prove the following auxiliary result: consider a sequence g of L p () so that g L p () < +, then there exists G L p () so that n = g G pointwise almost everywhere and in L p (). Proof of the auxiliary result: Define h n := n = g and h := = g two functions from to [0, + ]. Since h p n increases and converges pointwise to h p in [0, + ], the Beppo-Levi theorem implies hp n dµ hp dµ. Moreover by triangular inequality h n Lp () n = g Lp () M, and thus h L p () and is finite almost everywhere. Hence, the series g converges absolutely almost everywhere, hence converges almost everywhere, and we call this pointwise limit G. This limit satisfies G(x) = g = g (x) = h(x) and is therefore L p. Finally the integral n G(x) g (x) = p dµ(x) 0 by applying the dominated convergence theorem: the integrand converges pointwise to zero, and the following domination holds: G(x) n = g (x) p 2 p h p, where h p integrable. This concludes the proof of the auxiliary result. Bac to the main proof: consider f n Cauchy sequence, we choose a subsequence f ϕ() so that the series g := f ϕ(+) f ϕ() satisfies g p, which implies that 2 = g L p () < +. The auxiliary result shows that there is a G L p () so that n = g = f ϕ(n+) f ϕ() G pointwise almost everywhere and in L p (), and we define f := G + f ϕ(). Then (Cauchy property) f m f n p dµ 0 as m, n and taing m = ϕ() we get f n f in L p. When p = + : by removing the null set A := m,n A m,n with A m,n := {x f m (x) f n (x) > f n f m, we are left with a sequence uniformly Cauchy for on \ A and the rest of the proof is standard. Remar.23. Observe that we have proved that the convergence in L p implies the pointwise convergence almost everywhere of a subsequence. (Is it always true for the complete sequence?) Theorem.24 (Abstract density result). Consider (, A, µ) a measured space and p [, + ]. Then the simple functions that belong to L p () are dense in L p (). Remar.25. Note that a simple function s = n i= α iχ Ai belongs to L p (p [, + )) iff µ(a i ) finite as soon as α i 0, and all simple functions belong to L. Proof. For f real or complex, split real/imagnary positive/negative parts and approximate each: it is enough to deal with non-negative functions. For f 0, use again s n so that s n (i )/2 n on A i n and s n n on B n with B n := {x f(x) > n} and A i n := {x f(x) [(i )/2 n, i/2 n ]} for i =,..., n2 n. This produces a sequence that converges pointwise and from below towards f 0. When p = + one checs that the convergence is uniform, and when p [, + ) the Beppo-Levi theorem then implies the convergence in L p norm. Theorem.26 (Density-separability result in R n ). Consider O open set of R n and p [, + ), then () L p (O) is separable (i.e. has a countable dense subset), (2) smooth functions compactly supported in O are dense in L p (O). We will need and admit a ey property of the Lebesgue measure: Theorem (Regularity of the Lebesgue mesure) A regular measure on a topological space with a σ-algebra of measurable sets is a measure for which every measurable set can be approximated from above by measurable open sets (outer regularity) and from below by measurable compact sets (inner regularity). The Lebesgue measure on R n is regular for the Lebesgue sets. xercise. Observe that it implies that any Lebesgue set of R n with finite measure is squeezed between two Borel sets with same measure. Proof. Let us prove () separability: Consider the countable base C of open sets of R n made up of hypercubes with rational coordinates.

6 6 C. MOUHOT Observe that: any open set O can be covered with a countable union of rational cubes with disjoint interiors. It is proved by an inductive construction: on the grid Z n retains the cubes fully inside O, discard those fully outside, for those remaining bisect them into 2 n smaller half-length cubes and iterate. Then we approximate any simple functions n i= α iχ Ai by simple functions using elements C and rational coefficients as follows: the outer regularity of the measure allows approximation of A i by an open set O i (with small error on the measure), then we cover O i with a countable collection of closed cubes C i, as above, with µ(o i ) = = µ(c i,), and since the series converges the partial sum N i = µ(c i,) can approximate the target measure with small error, with a finite number of those cubes; we finally approximate the coefficient with a rational number. Point (2) of the statement: we construct smooth compactly supported approximation of the characteristic function on each cube. The approximation of the characteristic function by a continuous function is straightforward with a piecewise affine function. The approximation by a smooth compactly function requires the use the function e /x2 as seen in Analysis II. Remar.27. Approximation by the convolution with an approximation of the unit requires anyway the construction of a smooth compactly supported function, using e /x2 or a variant. Then the convergence ϕ f f in L p for ϕ approximation of the unit relies on the continuity of the translation operator in L p, p [, + ). Remar.28. Various generalisations of point () of the previous statement are possible, e.g. secondcountable with a regular measure. xercise 2. Prove that L (R n ) is not separable (hint: consider the family χ Br(0) for r > 0)..4. How regular are measurable and integrable functions? Definition.29 (Lesbegue points). For a function f : R n C we say that x is a Lebesgue point if f(y) f(x) dµ(y) 0 (averaging limit). B r(x) B r(x) xercise 3. Prove that all points of continuity are Lebesgue. Theorem.30 (Lebesgue s differentation and density Theorems). For an integrable function f L (R n ), almost every points are Lebesgue (differentiation). This implies on the Borel sets: let B(R n ) then for almost every x R n the density ratio Br(x) B r(x) χ (x) as r 0 (density). Proof of the density theorem assuming the differentiation theorem. For x M, r <, the ratio rewrites B r(x) B r(x) = BM+(x) Br(x) B r(x), then apply the differentiation theorem to f = χ BM+(x). Proof of the differentiation theorem. General strategy: The desired limit is satisfied for continuous functions and L functions can be approximated by continous functions in L, hence it is enough to prove that the amount of non-lebesgue points is controlled by the L norm of the function. This amount will be measured by the following operator. Step : Define the Hardy-Littlewood maximal operator Mf(x) := sup r>0 mf(x, r) with mf(x, r) := µ(b(x,r)) B(x,r) f dµ. Then for a > 0, a := {x Mf(x) > a} is an open set. ( ) n r If x a, then Mf(x) > a and there is r > 0 s.t. mf(x, r) > a. Let ɛ > 0 s.t. r+ɛ > a mf(x,r). Then for y B(x, ɛ), ( ) n ( ) n r r mf(y, r+ɛ) = f dµ f dµ = mf(x, r) > a µ(b(y, r + ɛ)) B(y,r+ɛ) r + ɛ µ(b(x, r)) B(x,r) r + ɛ where we have used B(x, r) B(y, r + ɛ). This proves the step. Remar it implies that M f : R n [0, + ] is measurable, since the sets (a, + ] generate the Borel sets. Step 2: Vitali s covering lemma: Consider a set R n that is included in a finite number of open balls N i= B(x i, r i ). There a subset of indeces J {,..., N} s.t. the collection of balls (B(x j, r j )) j J are pairwise disjoint and j J B(x j, 3r j ). Assume wlog the radiuses are raned r r 2 r N. Then consider B(x, r ): all balls that intersect it are included in B(x, 3r ), remove them, then consider the ball with the second largest radius in this new collection and argue similarly, and continue inductively. By finite induction it builds the desired collection.

7 ANALYSIS OF FUNCTIONS 7 Step 3: For all a > 0 one has µ( a ) 3n a f L (R n ). Consider any compact set K a : for any x K a there is r x so that mf(x, r x ) > a. Then K x K B(x, r x ) N i= B(x i, r xi ) (finite covering by compactness). The Vitali covering Lemma gives then indeces J s.t. K J B(x j, 3r xj ) with disjoint balls B(x j, r xj ), j J. We deduce µ(k) µ(b(x j, 3r xj )) = 3 n µ(b(x j, r xj )) 3n f dµ 3n f dµ. a j J j J j J B(x j,r xj ) a R n Since (inner regularity) µ( a ) = sup{µ(k) K a compact}, it concludes this step. Step 4: Conclusion. Define the operators tf(x, r) = µ(b(x,r)) B(x,r) f(x) f(y) dµ(y) and T f(x) = sup r>0 t f (x, r). Suppose f = g + h with g L (R n ) continuous and h L (R n ). Then tf(x, r) t g (x, r) + h(y) dµ(y) + h(x) µ(b(x, r)) and taing r 0 and using that all points are Lebesgue for a continuous function: T f(x) Mh(x) + h(x). Therefore ({ µ x T f(x) > }) ({ µ x Mh(x) > }) ({ + µ x h(x) > }) 2(3 n + ) h L 2 2 (R n ) where we have used Step 2 on Mh and Chebychef s inequality. The density of continuous functions in L then mean the decomposition f = g+h exists with h L as small as wanted, which implies µ({x T f (x) > }) = 0 for all non-zero integer, and, by union over, so µ({x T f (x) 0}) = 0. Let us explore further the lins between integrability and differentiability, as emphasized by the name of the previous theorem. Theorem.3. Consider f L (R) and define F (x) = x f(y) dµ, then F is differentiable almost everywhere with F = f (understood as an equality between elements of L (R), i.e. almost everywhere). F (x+δ) F (x) Proof. Observe that f(y) dµ. Then µ([x, x + δ]) δ = µ([x,x+δ] [x,x+δ] [x,x+δ] f(y) f(x) dµ B(x,r) 2 µ(b(x, δ)) B(x,δ) f(y) f(x) dµ which goes to zero for almost every x R by the previous theorem. Hence for almost every x R, one has F (x+δ) F (x) δ f(x), which concludes the proof. xercise 4. Is the converse of this result true? More precisely if f differentiable almost everywhere with f L (R), do we always have f(y) f(x) = y x f (z) dµ? Remar.32. One can prove that being the integral of an L function (i.e. f(y) f(x) = y x f (z) dz with f L ) is equivalent to the absolute continuity: for any ε > 0 there is δ > 0 s.t. for any finite collection of pairwise disjoint intervals (a, b ), =,..., n with n = (b a ) δ, then n = f(b ) f(a ) ε. We finally turn to the lin between pointwise and uniform convergence, and between measurability and continuity. Theorem.33 (gorov). Let f : R n C be a sequence of measurable functions, a Borel set A with finite measure and assume that f converges pointwise to a function f on A. Then for any ε > 0 there is a Borel set A ε A with µ(a \ A ε ) ε s.t. f n converges uniformly to f on A. Proof. Define for, l the sets (l) := p {x A f p (x) f(x) l }, prove for any, l : (l) (l) +, (l+) (l) and prove for any l : A = (l). At l fixed, A = (l) with nondecreasing union so (continuity from below of measure) there is l so that Δ l := A \ (l) l has measure µ(δ l ) ε/2 l. Define Δ = l Δ l which has measure less than ε, and A ε := A \ Δ. On A ε the convergence is uniform: for any l, A ε () l and thus sup x Aε f p (x) f(x) / for all p l. For those interested in digging more on the web, this means that Mf is wea L with wea L norm bounded in terms of the L norm of f. This is one example of Hardy-Littlewood maximal inequalities.

8 8 C. MOUHOT xercise 5. Prove that the assumption µ(a) < + is necessary for the result to hold. Theorem.34 (Lusin s Theorem - tae ). If f : R C is measurable and ε > 0, there is some Lebesgue measurable set R with µ() < ε so that f R\ is continuous. Proof. Let us first present a proof which shows an interesting approximation by step functions. Step. Given a measurable set F R with µ(f ) < and ε > 0, there exists a finite union K of intervals with µ(f ΔK) < ε. We have already proved it when studying the separability of L (R). Step 2. Given a simple function s on a measurable set F with finite measure and ε > 0, there is a step function S so that µ({x R S(x) s(x)}) < ε. Let s = n = α χ A with A F, let K be a finite union of intervals with µ(a ΔK ) < ε/n, and let S = n = α χ K. Then S(x) = s(x) on F \ ( n = (A ΔK )), and µ( n = (A ΔK )) < ε. Step 3. If f : F C measurable with F measurable set with finite measure, there is a sequence of step functions converging to f almost everywhere. Tae a sequence of simple functions s n that converges pointwise to f and for each n tae S n step function so that S n = s n except on a set D n with measure µ(d n ) < 2 n. Then D := N n N D n has zero measure and S n f on F \ D. Step 4. Conclusion. Apply the previous step to each F l = [l, l + ) to get sequences of step functions Sn l converging to f almost everywhere on F l. gorov s Theorem shows that there is A l F l with µ(a l ) ε 3 2 l s.t. the convergence Sn l f is uniform in F l \ A l. Let A := l Z A l, which has measure µ(a) ε. The set D of points of discontinuity of all step functions on all intervals is countable, hence := Z D A has measure µ() ε. The function S n concatenating all Sn l is continuous on R \ and converges uniformly to f on R \, which concludes the proof. An alternative short proof. It is enough to consider f : F R with µ(f ) < +, by applying on each F l and to the real or imaginary parts of f. Let (V n ) n be an enumeration of the open intervals with rational endpoints in R. Fix compact sets K n f (V n ) and K n F \ f (V n ) for each n so that µ(f \ (K n K n)) < ε/2 n (inner regularity). Fix open sets U n s.t. K n U n and U n K n =. Now, for K := n (K n K n), µ(f \ K) < ε. Given x K and an n with f(x) V n, x K n U n and f(u n K) V n. Since the V n are a base of neighbourhoods, this shows that f K is continuous. Remar.35. This theorem does not claim that f is continuous at every x R \. It is the restriction of f that is continuous. To illustrate the difference, consider f = χ Q, which is nowhere continuous. However, its restriction to R \ Q is continuous (constantly zero). Theorem.36 (Lusin s Theorem - tae 2). If f : R C is measurable and ε > 0, there is some measurable set G R with µ(g) < ε and a continuous function g : R C so that f = g on R \ G. Proof. Apply the previous theorem: Tae with µ() < ε/2 s.t. f R\ is continuous. Tae (outer regularity) G R open set that includes and s.t. µ(g) ε. This set G is a pairwise disjoint countable union of open intervals G = (a, b ). Finally define g by g := f on R \ G, and g(x) := f(a ) + x a b a ( f(b ) f(a )) on (a, b ). xercise 6. The full power of Lusin s theorem is not required to prove that continuous functions are dense in L (R). Give however a new proof of it by using Lusin s theorem.