Re: Measurable Functions (2)

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

Re: Measurable Functions (2)

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

Thanks anyway.