Measurable Functions (2)

**Markeur** · September 25th 2011, 10:08 AM

Let $E$ be a measurable subset of $\mathbb{R}$ with $m(E) < \infty$ and let $f$ be a real-valued measurable function on $E$ . Prove that for any $\epsilon > 0$ , there exists a bounded measurable function $g$ on $E$ such that $m(\{x \in E : f(x) \not= g(x)\}) < \epsilon$ .

Again, how do I go about doing it? Thanks in advance.

**Tinyboss** · September 25th 2011, 01:35 PM

Okay, so what's the main issue here? If you just let $g:=f$ , then the measure of the set where they differ is certainly less than $\epsilon$ . But of course, then $g$ may not be bounded as required. So try defining $g_n(x):=\mathrm{min}(f(x),n)$ , and looking at what happens to $m(\{x\mid f(x)\ne g_n(x)\})$ as $n$ goes to infinity.

**Markeur** · September 26th 2011, 06:39 AM

Okay managed to solved it. Though I used g(x) = f(x) if x lies in A, g(x) = 0 otherwise.

Thanks anyway.

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