# Math Help - Measurable Functions (2)

1. ## Measurable Functions (2)

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 $m(\{x \in E : f(x) \not= g(x)\}) < \epsilon$.

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.