My Professor stated the following claim without proof:
If $f$ is measurable, then for all $\epsilon > 0$, there exists $E$ and a simple function $\phi$ such that $\mu(E) < \epsilon$ and $|f - \phi| < \epsilon$ on $E^{C}$.
I claim to have a counter-example:
Take $(\mathbb{R}, \mathscr{L}, m)$ and $f = |x|$ (which is continuous and, hence, measurable).
Let $\epsilon = 1/2$. Take any measurable set $E$ satisfying $m(E) < 1/2$. Then $\mathbb{R} \setminus E$ is unbounded (if there were $M$ bounding $\mathbb{R} \setminus E$, then $E$ would contain $\{x > M\}$ and $\{x < -M\}$). Now choose any simple function $\phi$. By definition $\phi$ takes on an finitely many values. Thus $\phi$ cannot approximate $f = |x|$ to $\epsilon$ precision, since $f$ takes on infinitely many values on the unbounded set $\mathbb{R} \setminus E$.
Is my counter-example valid?## Heading ## Perhaps this claim is true if $\mu(X) < \infty$?
I think finite measure is needed, then it amounts to the Egorov Theorem:
Given a sequence $(\phi_{n})$ of simple functions such that $\phi_{n}\rightarrow f$ a.e. on $D$, $|D|<\infty$, then $\phi_{n}\rightarrow f$ almost uniformly.