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Measure, Integration & Real Analysis, 2021a

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Section 2E Convergence of Measurable Functions 65<br />

Approximation by Simple Functions<br />

2.88 Definition simple function<br />

A function is called simple if it takes on only finitely many values.<br />

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

c 1 ,...,c n are the distinct nonzero values of f . Then<br />

f = c 1 χ<br />

E1<br />

+ ···+ c n χ<br />

En<br />

,<br />

where E k = f −1 ({c k }). Thus this function f is an S-measurable function if and<br />

only if E 1 ,...,E n ∈S(as you should verify).<br />

2.89 approximation by simple functions<br />

Suppose (X, S) is a measurable space and f : X → [−∞, ∞] is S-measurable.<br />

Then there exists a sequence f 1 , f 2 ,...of functions from X to R such that<br />

(a) each f k is a simple S-measurable function;<br />

(b) | f k (x)| ≤|f k+1 (x)| ≤|f (x)| for all k ∈ Z + and all x ∈ X;<br />

(c)<br />

(d)<br />

lim<br />

k→∞<br />

f k (x) = f (x) for every x ∈ X;<br />

f 1 , f 2 ,...converges uniformly on X to f if f is bounded.<br />

Proof The idea of the proof is that for each k ∈ Z + and n ∈ Z, the interval<br />

[n, n + 1) is divided into 2 k equally sized half-open subintervals. If f (x) ∈ [0, k],<br />

we define f k (x) to be the left endpoint of the subinterval into which f (x) falls; if<br />

f (x) ∈ [−k,0), we define f k (x) to be the right endpoint of the subinterval into<br />

which f (x) falls; and if | f (x)| > k, we define f k (x) to be ±k. Specifically, let<br />

⎧<br />

m<br />

if 0 ≤ f (x) ≤ k and m ∈ Z is such that f (x) ∈ [ m<br />

, m+1 )<br />

2 k 2 k 2 ,<br />

k<br />

⎪⎨ m+1<br />

if − k ≤ f (x) < 0 and m ∈ Z is such that f (x) ∈ [ m<br />

, m+1 )<br />

2<br />

f k (x) =<br />

k 2 k 2 ,<br />

k<br />

k if f (x) > k,<br />

⎪⎩<br />

−k if f (x) < −k.<br />

Each f −1( [ m 2 k , m+1<br />

2 k ) ) ∈Sbecause f is an S-measurable function. Thus each f k<br />

is an S-measurable simple function; in other words, (a) holds.<br />

Also, (b) holds because of how we have defined f k .<br />

The definition of f k implies that<br />

2.90 | f k (x) − f (x)| ≤ 1 2 k for all x ∈ X such that f (x) ∈ [−k, k].<br />

Thus we see that (c) holds.<br />

Finally, 2.90 shows that (d) holds.<br />

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

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