integration - for each $epsilon >0$ there is a $delta >0$ such that whenever $m(A)

This is an old preliminary exam problem:

Show that, for every nonnegative Lebesgue integrable function $f:[0,1]\rightarrow \mathbb{R}$ and every $\epsilon>0$ there exists a $\delta>0$ such that for each measurable set $A\subset [0,1]$ with $m(A)<\delta$ it follows that $\int_A f(x)dx<\epsilon$.

Here's my attempt at a proof: Since $[0,1]$ is compact, and $f$ is real-valued, there exists an $M>0$ such that $f(x)\le M$ for all $x\in [0,1]$. Therefore, for $\epsilon>0$, let $\delta=\epsilon/M$. Then for all $A\subset [0,1]$ such that $m(A)<\delta$, we have that $\int_A f(x)dx\le Mm(A)<\epsilon$. Where here $m$ denotes Lebesgue measure.

The part I'm unsure about is the existence of $M$. If the function is continuous, then there is no problem, but $f$ does not have to be continuous to be Lebesgue measurable. On the other hand, the problem says that $f$ is real-valued, not extended real-valued, so this means that $f(x)$ is defined and finite for each $x$, right?

Answer

Any integrable $f:[0,1] \to R$ can be approximated by a continuous function $f_\epsilon$ on $[0,1]$ up to $\epsilon$ in $L^1$. With this, the fix to your argument is to use the triangle inequality:

$$\left| \int_A f dx \right| \leq \left| \int_A (f-f_\epsilon) dx \right| + \left| \int_A f_\epsilon dx \right| \leq \epsilon + \delta M.$$

Epsilon is picked first, from this we get an $M$ dependent on $f_\epsilon$ (dependent on $\epsilon$) and from this we can pick $\delta = \epsilon/M$. to get the upper bound $2 \epsilon$.

If you don't feel comfortable with a continuous approximation (ala Lusin's theorem) you can use simple functions instead.