An integral of a positive function over a "small" set is "small".

Question

Consider a measure space $(X,\mathfrak{A},\mu)$, $f$ an integrable function and $f \geq 0$. Show that for every $\epsilon > 0$ there exists a $\delta > 0$ such that for every $A \in \mathfrak{A}$ if $\mu(A)<\delta$ then $\int_A f d\mu < \epsilon$.

I am given a hint which is to consider $f_n(x) = \min\{n, f(x)\}$.

I do this and given that $f_n \leq f_{n+1}$ I can use the Beppo-Levi theorem and conclude that $\displaystyle \int_A f d\mu = \lim_{n \to \infty} \int_A f_n d\mu$.

If $f$ is bounded I can further conclude that there exists M such that $\displaystyle\lim_{n \to \infty} \int_A f_n d\mu \leq \int M d\mu = M\mu(A)$ and take $\delta = \frac{\epsilon}{M}$.

I don't really know what to do if $f$ is not bounded.

Answer 1

Suppose the result were false. Then there would exist $\epsilon>0$ and a sequence $(A_n)$ of measurable sets such that $\mu(A_n)<2^{-n}$, and $\int_{A_n}fd\mu \geq \epsilon$. By the Borel-Cantelli lemma $\mu(\limsup_n A_n)=0$. Let $A = \limsup_n A_n$. Reverse Fatou's lemma now implies $$0=\int_A f d\mu \geq \limsup_n \int_{A_n}fd\mu \geq \epsilon.$$ This is a contradiction. (Note that the argument works for all integrable $f$ by replacing $f$ with $|f|$.)

Answer 2

Whether $f$ is bounded or not, $f$ being integrable implies that $f_n\to f$ in $L^1$. Thus, fix some positive $\epsilon$, and choose some $n$ such that $\|f_n-f\|_1\leqslant\epsilon$.

Now, for every measurable $A$, $\left|\int_Afd\mu\right|\leqslant\left|\int_Af_nd\mu\right|+\|f_n-f\|_1\leqslant n\mu(A)+\epsilon$, hence, for every measurable $A$ such that $\mu(A)\leqslant\epsilon/n$, one gets $\left|\int_Afd\mu\right|\leqslant2\epsilon$, as desired.

One sees that this approach works for every integrable function $f$ with real values, not only nonnegative, simply replacing $f_n=\min\{f,n\}$ by $f_n=\max\{-n,\min\{f,n\}\}$.