prove the equivalence between a null set and a limit

Question

I'm asked to prove that for any non-negative, measurable and integrable function $f$ on $[0,1]$, we have $\lim\limits_{a\to 0}\int_{0}^{a}fdx=0$. I want to use the theorem that for null set E, such a function has $\int_E f=0$. And I construct $E=\bigcap_{n=1}^{\infty}[0,1/n]$. And I claim $\int_E f=0$ gives $\lim\limits_{a\to 0}\int_{0}^{a}fdx=0$. Is it correct? If not, how can I fix this? Or whether there are other ways to do it?

Answer 1

Use the Dominated convergence theorem. Let $(x_k)$ be any sequence decreasing to zero (i.e, $x_k \downarrow 0$). Then $$\lim_{k \to \infty} \int_0^{x_k} f(x) dx = \lim_{k \to \infty} \int_0^1 f(x)\; 1_{[0,x_k]}(x)\; dx = \int_0^1 \bigg[\lim_{k \to \infty} f(x)1_{[0,x_k]}(x) \bigg]dx = \int_0^1 0 \; dx=0$$ We are allowed to apply DCT because all of the functions $f\cdot1_{[0,x_k]}$ are bounded above by the integrable function $f$.

We showed that $\lim_{k \to \infty} \int_0^{x_k} f(x) dx=0$ whenever $x_k \downarrow 0$, which shows that $\lim_{a\to 0} \int_0^a f \;dm=0$.

I just want to make an additional remark, that this problem is the special case of a more general result regarding absolute continuity: namely, if $f\in L^1(\mu)$, then for any $\epsilon>0$, there exists $\delta >0$ such that $\mu(E)<\delta$ implies $\big| \int_E f\; d\mu\big| < \epsilon$.