Measurable Functions

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Dennis Gardner
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1 Measurable Fuctios Dug Le 1 1 Defiitio It is ecessary to determie the class of fuctios that will be cosidered for the Lebesgue itegratio. We wat to guaratee that the sets which arise whe workig with these fuctios are measurable. I this cotext, as well as i may others, the iverse images of sets are more useful tha the images of sets. For example, a fuctio is cotiuous if ad oly if the iverse image of each ope set is ope. A variatio of this leads to the cocept of measurable fuctio. Defiitio 1.1 A fuctio f : E IR is measurable if E is a measurable set ad for each real umber r, the set {x E : f(x > r} is measurable. As stated i the defiitio, the domai of a measurable fuctio must be a measurable set. I fact, we will always assume that the domai of a fuctio (measurable or ot is a measurable set uless explicitly metioed otherwise. From the defiitio, it is clear that cotiuous fuctios ad mootoe fuctios are measurable. However, just as there are sets that are ot measurable, there are fuctios that are ot measurable. Let E be a measurable set with positive measure ad let A E. represets the characteristic fuctio of A. That is, χ A (x = { 1 x A 0 x A The fuctio χ A The set {x E : χ A (x > r} is either, A, or E (check this!. So, χ A is a measurable fuctio if ad oly if A is a measurable set. Theorem 1.2 Let E be a measurable set ad let f : E IR. The followig are equivalet: (1 For each real umber r, the set {x E : f(x > r} is measurable. (2 For each real umber r, the set {x E : f(x r} is measurable. (3 For each real umber r, the set {x E : f(x < r} is measurable. (4 For each real umber r, the set {x E : f(x r} is measurable. Proof: To see (1 (2 we simply ote that (check it! {x E : f(x r} = {x E : f(x > r 1 }, =1 1 Departmet of Applied Mathematics, Uiversity of Texas at Sa Atoio, 6900 North Loop 1604 West, Sa Atoio, TX dle@math.utsa.edu 1

2 {x E : f(x > r} = {x E : f(x r + 1 }. =1 The other implicatios are established by takig complemets. Whe workig with fuctios, sets of measure zero ca ofte be igored. A property is said to hold almost everywhere (we will abbreviate it by a.e. if it holds everywhere except for a set of measure zero. For example, the fuctios f : E IR ad g : E IR are equal almost everywhere if ad oly if the set {x E : f(x g(x} has measure zero. Similarly, f is cotiuous a.e. iff the set of discotiuity poits has measure 0. Theorem 1.3 Let f : E IR be measurable ad let g : E IR. If f = g a.e.o E, the g is measurable. Proof: For r IR, let A = {x E : f(x > r} ad A = {x E : g(x > r}. The A is measurable ad A B, B A are subsets of {x E : f(x g(x} so that they have zero measures. Sice B = (B A (B A = (B A (A (A B is measurable. Thus, g is measurable. 2 Operatios o measurable fuctios Theorem 2.1 Let f : E IR ad g : E IR be measurable fuctios ad let k IR. The the fuctios kf, f + g, f ad fg are measurable. I additio, the fuctio f/g is measurable if g(x 0 for all x E. Proof: We will oly prove that f +g is measurable as the others will be left as exercises. Let r be ay real umber. If f(x + g(x > r, the there exists a ratioal umber q such that r g(x < q < f(x. It follows that {x E : (f + g(x > r} = {x E : f(x > q} {x E : g(x] > r q}. q Q Hece, f + g is a measurable fuctio. The followig more geeral theorem will be useful. Theorem 2.2 Let f ad g be measurable fuctios defied o E. Let F be real ad cotiuous fuctio o IR 2. The the fuctio h(x = F (f(x, g(x is measurable. Proof: Let G r = {(u, v : F (u, v > r}. The G r is a ope subset of IR 2, ad thus we ca write it as a coutable uio of ope rectagles. That is, G r = I where I = {(u, v : a < u < b, c < v < d }. 2

3 The sets {x E : a < f(x < b } ad {x E : c < g(x < d } are measurable. Thus {x E : (f(x, g(x I } = {x E : a < f(x < b } {x E : c < g(x < d } is also measurable. Hece the same is true for {x E : h(x > r} = {x E : (f(x, g(x G r } = {x E : (f(x, g(x I }. Thus, h is measurable. The followig theorem shows that may fuctios we have leared i Real Aalysis I are measurable. Theorem 2.3 Let E be measurable. If f : E IR is cotiuous a.e., the f is measurable. Proof: Let D be the set of discotiuities of f. The µ(d = 0 ad all of its subsets are measurable. Let r IR ad ote that {x E : f(x > r} = {x E D : f(x > r} {x D : f(x > r}. We eed oly show that C = {x E D : f(x > r} is measurable (why?. For each x C, as f is cotiuous at x, we ca fid δ x > 0 such that if y V δx (x the f(y > r (why?. It is clear that C = (E D V δx (x, x C which is measurable (why?. This shows that f is measurable. Let {f } be a sequece of fuctios defied o E. We will deote sup f (x = sup{f (x : N} ad lim sup It is easy to check (do it! that lim sup Similarly, we defie f (x = lim (sup f k (x k if f (x = if{f (x : N} ad lim if It is easy to verify the followig relatios if f (x = sup( f (x ad lim if = if f (x = lim (sup f k (x k ( sup f k (x k.. (2.1 ( f (x = lim if f k(x. (2.2 k f (x = lim sup( f (x. (2.3 3

4 Theorem 2.4 Let {f } be a sequece of measurable fuctios defied o E. For x E, set g(x = sup The g ad h are measurable. f (x ad h(x = lim sup f (x. Proof: We eed oly to show that g is measurable sice the measurability of h comes from this fact ad the relatios (2.1, (2.3 (exercise!. For ay r IR, we have {x E : g(x > r} = {x E : f (x > r}. This shows that g is measurable. Let {f } be a sequece of fuctios o E that coverges a.e.to a fuctio f. This is to say that the set C = {x E : lim f (x exists} satisfies µ(e C = 0. Thus, f is oly defied o C. However, sice E C has measure zero, it does ot matter how the fuctios f are defied there! As this situatio will arise frequetly, we will agree upo the covetio that, uless otherwise stated, such fuctios are equal to zero o E C. Hece, f is ow coveietly defied o the whole set E! We ow have Corollary 2.5 Let f be a sequece be a sequece of measurable fuctios defied o E ad f : E IR. If {f } coverges poitwise to f a.e.o E, the f is measurable. The followig theorem provides aother useful characterizatio of measurable fuctios. Theorem 2.6 f : E IR is measurable iff f 1 (B is measurable for every Borel set B. Proof: Suppose that f is measurable. Let s defie the followig collectio A = {A IR : f 1 (A is measurable}. We ca easily check the followigs (exercise: A. If A A the A C A (f 1 (A C = f 1 (IR f 1 (A = E f 1 (A. If {A } is a sequece of sets i A the A A (f 1 ( A = f 1 (A. This shows that A is a σ-algebra. Moreover, sice f is measurable, ay ope iterval (a, b must be i A (That is, f 1 ((a, b is measurable (why?. Thus, A cotais all ope sets (why?. From the defiitio of Borel sets, we see that they must be i A. Hece, f 1 (B is measurable for all B B. The coverse is trivial (check it!. 4

5 Exercises 1. Let f : E IR ad A IR. Show that f 1 (A C = E f 1 (A. Moreover, if {A α } is a collectio of sets the f 1 ( α A α = α f 1 (A α ad f 1 ( α A α = α f 1 (A α. 2. Let A IR. Determie r such that {x E : χ A (x > r} is either (respectively, A, or E. 3. Prove the set equalities i the proof of Theorem If f is measurable, the show that f 1 ((a, b is measurable for ay a, b IR. 5. Complete the proof of Theorem 2.1. You caot use Theorem 2.2! 6. Which of the followig fuctios are measurable. Explai why. f(x = { { 1 x Q x 0 x Q, g(x = 2 x Q 0 x Q, h(x = 7. For each N, cosider the fuctio { 1 x x [0, 1/] f (x = 0 x > 1/ Let f(x = lim f (x. (a Is f cotiuous? (b Is f measurable? { x 3 + x 2 x Q si(x x Q 8. Let f : IR IR. Show that f is cotiuous iff f 1 (A is ope for every ope set A. 9. Prove (2.1 ad ( Show that the fuctio h defied i Theorem 2.4 is measurable. 5

SAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx

SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval