9 Measurable functions and their properties Throughout this section we assume that (X, ) is a measurable space. That is, we assumeX is a non-empty set and is aσ-algebraM inX. The sets in are called measurable subsets ofX. M M Notation (range, pre-image): SupposeY is a set andf:X Y is a function. The range off, denotedf(X), is the set of possible values taken by→ the functionf, i.e. f(X) := f(x):x X . { ∈ } GivenB Y , we define the pre-image (or inverse image) ofB underf as ⊂ 1 f − (B) := x X:f(x) B . { ∈ ∈ } Note that nothing is implied aboutf having an inverse function: in general, it will not 1 1 have one. For example,f − (B) could be empty, even ifB is not. The notationf − (B) is simply a shorthand for the set of points inX thatf maps into the setB. For example if 2 1 1 X=Y=R andf(x) =x thenf − ([1, 9]) = [1, 3] [ 3, 1] whilef − (( , 0)) =∅. ∪ − − −∞ Notation: Extended real line. We are mainly interested in functionsf fromX toR but sometimes it is useful to allowf to take values as well. We write R or ±∞ [ , ] to denote the setR ,+ , known as the extended real line. −∞ ∞ ∪{−∞ ∞} 1 Definition 9.1. We say that a functionf:X R is measurable, iff − ((α, ]) → ∞ ∈M for everyα R. ∈ If in factf(X) R, then we still use the same definition of measurability off; in this 1 ⊂ 1 case the setf − ((α, ]) will be the same asf − ((α, )). ∞ ∞ Remark. If we wish to clarify whichσ-algebra inX we mean (for example if more than oneσ-algebra is being considered inX, a situation common in probability theory), then we sayf is measurable with respect to (or for short, -measurable). M M Example 9.2(Indicator functions). LetE X. The function1 E :X R, defined by ⊂ → 1 ifx E; 1E(x) := ∈ 0 ifx E, � �∈ is called the indicator function or characteristic function ofE. The function1 E is a measurable function, if and only ifE (HW). ∈M Definition 9.3. SupposeW R is Borel (the setW could be all ofR), and let ⊂ := B W:B . BW { ⊂ ∈B} Then (W, W ) is a measurable space. A functiong:W R is called a Borel measur- able functionB , or just Borel function, if it is measurable→ with respect to . B W Lemma 9.4. (‘All continuous functions are Borel’). Suppose thatW R is open, ⊂ and thatf:W R is continuous. Thenf is Borel measurable on(W, W ). → B 28 Proof. Recall the continuity off is defined as follows: for allx W and allε> 0, there existsδ> 0 such that whenever∈ y x <δ then f(y) f(x) <ε. | − | | − | Letα R. Ifx W withf(x)>α, then we canfindδ> 0 such that for y x <δ we havey∈ W and∈f(y)>α. | − | ∈ Thereforef −((α, ]) =f −((α, )) W is open, and hence it is in W . Therefore,f is measurable on (W,∞ ). ∞ ⊂ B B W Lemma 9.5. SupposeY is a set andf:X Y is a function. Let := E Y: 1 → F { ⊂ f − (E) . Then is aσ-algebra inY. ∈M} F Proof. We leave this as HW. Theorem 9.6. Supposef:X R is a measurable function, andE is a Borel set inR. 1 → Thenf − (E) . ∈M 1 Proof. Set := E R:f − (E) . By Lemma 9.5, is aσ-algebra. Forα R F { ⊂ ∈M} F ∈ we have (α, ] by assumption, so that forα,β R withα<β we have that ∞ ∈F ∈ (α,β]=(α, ] (β, ] . Therefore (the bounded half-open intervals inR) so σ( )= ∞ by\ Theorem∞ ∈F 2.9. F⊃I F⊃ I B Theorem 9.7 (Composition of measurable functions). LetU R be an open set. If ⊂ f:X U is measurable, andg:U R is Borel (for example: if it is continuous), then → → h=g f, defined byh(x) =g(f(x)),h:X R, is measurable. ◦ → Proof. Letα R. We have: ∈ 1 h− ((α, ]) = x X:h(x)>α = x X:g(f(x))>α ∞ { ∈ } { ∈1 } = x X:f(x) g − ((α, ]) { ∈1 1 ∈ ∞ } =f − (g− ((α, ])). ∞ 1 Now,g − ((α, ]) sinceg is assumed Borel, and hence by Theorem 9.6, the set 1 1 ∞ ∈B f − (g− ((α, ])) is measurable (i.e., in ), as required. ∞ M Notation: Positive and negative parts. Fory R, we definey + := max(y, 0) ∈ (the positive part ofy) andy − := max( y, 0) (the negative part ofy). Note that + + − y=y y − and y =y +y −. Notation:− adding| | and multiplying functions. Suppose that we are given func- tionsf:X R,g:X R, and (forn N)f n :X R, and alsoa R. We → → + ∈ → ∈ define the functions f , f , f −, af, f+g andfg pointwise. That is, forx R we set | |+ + ∈ f (x) := f(x) , andf (x) := (f(x)) = max(f(x), 0), andf −(x) := max( f(x), 0), and (|af| )(x) :=| a(f|(x)), and (f+g)(x) :=f(x) +g(x) and (fg)(x) :=f(x)g(x). − Similarly we define the function supn fn pointwise, i.e. (supn fn)(x) = supn(fn(x)) for allx X. Likewise the functions lim sup n fn, infn fn, and lim infn fn and ∈ →∞ →∞ limn fn (if it exists) are defined pointwise, where we recall that for any sequence of →∞ numbers (an)n N we define ∈ lim sup an := lim (sup an, an+1, an+2,... ); n n { } →∞ →∞ lim inf an := lim (inf an, an+1, an+2,... ) n n →∞ →∞ { } 29 + + Remark 9.8. We note that f =f +f − andf=f f − (using pointwise addi- tion/subtraction both times).| | − + Corollary 9.9. Leta R. Iff:X R is measurable, so are af, f ,f andf −. ∈ → | | Proof. This follows from Theorem 9.7, takingg(y) = ay,g(y) = y ,g(y) =y + and | | g(y)=y − respectively. Theorem 9.10 (Sums and products of measurable functions). Iff:X R andg: → X R are measurable, then so aref+g andfg. → Proof. (ii) Supposef andg are measurable. Letα R. For any real u, v withu+v>α (so thatu>α v) there exists rationalq withu>q>∈ α v. Hence − − 1 (f+g) − ((α, ]) = x:f(x) +g(x)>α = q Q x:f(x)>q>α g(x) ∈ ∞ { } ∪ 1 { 1 − } = q Q[f − ((q, )) g − ((α q, ))] . ∪ ∈ ∞ ∩ − ∞ ∈M + Hencef+g is measurable. For the second part, note thatf=f f − so − + + + + + + fg=(f f −)(g g −) =f g +f −g− f g− g f −, − − − − so it suffices to provefg is measurable for the case withf 0 pointwise (i.e.f(X) [0, ) andg 0 pointwise. But in that case, forα> 0, iff(x)≥g(x)>α then there exists⊂ rational∞ q with≥ f(x)g(x)>q>α sof(x)>q>α/g(x), so x:f(x)g(x)>α = q Q (0, ) x:f(x)>q>α/g(x) , { } ∪ ∈ ∩ ∞ { }∈M and for the caseα = 0 we have x:f(x)g(x)>0 = n∞=1 x:f(x)g(x)>1/n , while forα< 0 we have x:f(x)g({x)>α =X }. ∪ { }∈M { } ∈M Remark 9.11. It can further be shown that iff:X [ , ] is measurable, then so + → −∞ ∞ aref := max f,0 andf − := max f,0 . Also, if f,g:X [0, ] are measurable, then so aref+g{ and}fg. (HW) {− } → ∞ Theorem 9.12 (Limits of measurable functions). Iff n :X R, are measurable func- → tions, defined forn N, then the functionsg:X R andh:X R defined by ∈ → → g := sup fn andh := lim sup fn n 1 n ≥ →∞ are also measurable. Similarly for infn 1 fn, and lim infn fn. ≥ →∞ Proof. For anyα R we have ∈ 1 g− ((α, ]) = x X : sup fn(x)>α ∞ { ∈ n 1 } ≥ = x X:f (x)>α for somen 1 { ∈ n ≥ } ∞ = x X:f (x)>α { ∈ n } n=1 � ∞ 1 = f − ((α, ]) . n ∞ ∈M n=1 , since � ∈M fn meas. � �� � 30 Thereforeg is measurable. Also g := inf n 1 fn = (sup n 1 fn) is also measurable. ≥ We can write − ≥ − h� = lim sup fk = inf sup fk, n k n n 1 k n →∞ ≥ ≥ ≥ and this is measurable by the previous paragraph. Similarly, lim infn fn = supn 1 infk n fk ≥ is also measurable. ≥ Corollary 9.13. Iff n :X R are measurable, andf(x) := lim n f(x) exists in R for →∞ eachx X, thenf is also measurable.→ ∈ For the proof of this, just note that iff(x) := lim n fn(x) exists, thenf = lim sup n fn. →∞ →∞ Definition 9.14. A functionf:X R is said to be simple, if (i) it is measurable and (ii) the rangef(X) is afinite set, i.e.→f takes onlyfinitely many values. Note 1: Here we exclude from the possible values. Note 2: It is convenient to±∞ include measurability in the definition of a simple function, though not all authors do so. Theorem 9.15 (Existence of Approximating Simple Functions). Letf:X [0, ] → ∞ be measurable. There exist nonnegative simple functionsf n,n N, such thatf n f pointwise, or in other words, such that for allx X: ∈ ↑ ∈ (a)0 f n(x) f n+1(x) for alln N; (b)f≤(x) f(x≤) asn . ∈ n → →∞ That is, every nonnegative measurable function can be expressed as an increasing limit of simple functions. Proof. For eachn 1, define the functionf n :X R by: ≥ → n n n n (k 1)10 − if (k 1)10 − f(x)<k10 − for some integer 1 k n10 ; fn(x) := − − ≤ ≤ ≤ n iff(x) n. � ≥ In other words,f n(x) is obtained by roundingf(x) down to thenth decimal place if f(x)<n, and settingf (x) =n iff(x) n.