Let $f$ be L measurable on [0,1] with $f(x)>0$ a.e. Suppose $A_{n}$ is a sequence of measurable sets in [0,1]. Prove that if $\lim_{n \to\infty} \int_{A_{n}}^{ } f(x) dx=0 $ then $\lim_{n\to\infty} \lambda{(A_{n})} =0 $
I am new to measure theory and couldn't find how to start. May I have a help,please?
Thank you
It suffices to show that $\limsup\limits_{n\rightarrow\infty}{\lambda(A_n)}<\epsilon$ for all $\epsilon>0$. To do so, fix $\epsilon>0$. Then there exists $\delta>0$ such that if $E_{\delta} = \{x\in[0,1]:f(x)\le\delta\}$, then $\lambda(E_{\delta})<\epsilon$ (why?). Now, let $B_n^{\delta} = A_n\backslash E_{\delta}$. Since $A_n\backslash B_n^{\delta}\subset E_{\delta}$, we have $\lambda(A_n\backslash B_n^{\delta})<\epsilon$. So it suffices to show that $\lambda(B_n^{\delta})\rightarrow 0$. Now $$\int\limits_{A_n}{f(x)\text{ d}x}\ge\int\limits_{B_n^{\delta}}{f(x)\text{ d}x}\ge\int\limits_{B_n^{\delta}}{\delta\text{ d}x}.$$ (Why is the last inequality true?) Can you conclude from here?