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Measure, Integration & Real Analysis, 2021a

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32 Chapter 2 <strong>Measure</strong>s<br />

The definition of an S-measurable<br />

function requires the inverse image<br />

of every Borel subset of R to be in<br />

S. The next result shows that to verify<br />

that a function is S-measurable,<br />

we can check the inverse images of<br />

a much smaller collection of subsets<br />

of R.<br />

Note that if f : X → R is a function and<br />

a ∈ R, then<br />

f −1( (a, ∞) ) = {x ∈ X : f (x) > a}.<br />

2.39 condition for measurable function<br />

Suppose (X, S) is a measurable space and f : X → R is a function such that<br />

f −1( (a, ∞) ) ∈S<br />

for all a ∈ R. Then f is an S-measurable function.<br />

Proof<br />

Let<br />

T = {A ⊂ R : f −1 (A) ∈S}.<br />

We want to show that every Borel subset of R is in T . To do this, we will first show<br />

that T is a σ-algebra on R.<br />

Certainly ∅ ∈T, because f −1 (∅) =∅ ∈S.<br />

If A ∈T, then f −1 (A) ∈S; hence<br />

f −1 (R \ A) =X \ f −1 (A) ∈S<br />

by 2.33(a), and thus R \ A ∈T. In other words, T is closed under complementation.<br />

If A 1 , A 2 ,...∈T, then f −1 (A 1 ), f −1 (A 2 ),...∈S; hence<br />

f −1( ∞ ⋃<br />

k=1<br />

A k<br />

)<br />

=<br />

∞⋃<br />

k=1<br />

f −1 (A k ) ∈S<br />

by 2.33(b), and thus ⋃ ∞<br />

k=1<br />

A k ∈T. In other words, T is closed under countable<br />

unions. Thus T is a σ-algebra on R.<br />

By hypothesis, T contains {(a, ∞) : a ∈ R}. Because T is closed under<br />

complementation, T also contains {(−∞, b] : b ∈ R}. Because the σ-algebra T is<br />

closed under finite intersections (by 2.25), we see that T contains {(a, b] : a, b ∈ R}.<br />

Because (a, b) = ⋃ ∞<br />

k=1<br />

(a, b − 1 k ] and (−∞, b) =⋃ ∞<br />

k=1<br />

(−k, b − 1 k<br />

] and T is closed<br />

under countable unions, we can conclude that T contains every open subset of R.<br />

Thus the σ-algebra T contains the smallest σ-algebra on R that contains all open<br />

subsets of R. In other words, T contains every Borel subset of R. Thus f is an<br />

S-measurable function.<br />

In the result above, we could replace the collection of sets {(a, ∞) : a ∈ R}<br />

by any collection of subsets of R such that the smallest σ-algebra containing that<br />

collection contains the Borel subsets of R. For specific examples of such collections<br />

of subsets of R, see Exercises 3–6.<br />

<strong>Measure</strong>, <strong>Integration</strong> & <strong>Real</strong> <strong>Analysis</strong>, by Sheldon Axler

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