Prove that to each $\epsilon >0$, there exists a $\delta >0$ so that the Lebesgue integral...

Suppose $f$ is in $L^1$ space of $\mu$, where $\mu$ is the Lebesgue measure. Prove that to each $\epsilon >0$, there exists a $\delta >0$ so that the Lebesgue integral of the absolute value of $f$ is less than $\epsilon$ (over the set $E$) whenever the measure of $E$ is less than $\delta$.

I know that the integral of $f$ is finite. So for arbitrary $f$ we can find the integral of $f$ to be less than some $\epsilon$, but how would I connect that with the measure of $E$ less than $\delta$. Any help would be welcome.

Solution 1:

The first thing you should understand is that "less than some $\epsilon$" will not cut it. You don't get to choose $\epsilon$ — it will be given to you, and it can be any positive number whatsoever. You must be prepared to produce $\delta>0$ (based on the $\epsilon$) such that the implication $$\mu(E)<\delta \implies \int_E |f|\,d\mu<\epsilon \tag{goal}$$ will hold.

Let's begin with something easy.

Easy case: $f$ is bounded, that is, there is a number $M$ such that $|f|\le M$ (almost everywhere). Then $$\int_E |f|\,d\mu \le \int_E M\,d\mu = M\,\mu(E) < M\delta \tag1$$ Since we want to conclude with "$\int_E |f|\,d\mu<\epsilon$", the right choice of $\delta$ is $\delta=\epsilon/M$. Now you are prepared for the challenge: no matter what $\epsilon$ someone throws at you, you choose $\delta=\epsilon/M$ and (goal) is achieved.

General case. If $f$ is not bounded, we can still introduce a bounded function into consideration, namely $|f|_M(x)=\min(|f(x)|,M)$ (here $M$ is ours to choose). This is called the truncation of $|f|$ at level $M$: if you don't see why, picture the graphs of $|f|$ and of $|f|_M$.

How to deal with $|f|_M$ is more or less clear (the easy case). What to do with the rest, namely $|f|-|f|_M$? Turns out that by choosing $M$ sufficiently large, we can make $\int_{\mathbb R} (|f|-|f|_M)\,d\mu$ is as small as we wish (I give a hint below). So, the strategy is:

• Pick $M$ so that $\int_{\mathbb R} (|f|-|f|_M)\,d\mu<\epsilon/2$.
• Pick $\delta = (\epsilon/2)/M$; this ensures that $\int_E |f|_M\,d\mu<\epsilon/2$.

Promised hint: since $|f(x)|=\lim_{M\to\infty} |f(x)|_M$ for almost every $x$, by the monotone convergence theorem $$\int_{\mathbb R} |f| = \lim_{M\to\infty}\int_{\mathbb R} |f|_M\,d\mu $$