For any pair of sequences $(a_n),(b_n)$ with $a_n<b_n$ and $a_n \rightarrow \infty$ we have $\lim_n \int_{a_n}^{b_n} f d \mu=0$

Question

I wrote down a solution, but I'm not sure if it works:

Question: Let $\mu$ be a measure on $\mathscr{B}(\mathbb{R})$ such that $\mu(B)< \infty$ for all bounded $B$. Let $f \geq 0$ be $\mu$-integrable. Show that for any pair of sequences $(a_n),(b_n)$ with $a_n<b_n$ and $a_n \rightarrow \infty$ we have $\lim_n \int_{a_n}^{b_n} f d \mu=0$.

Attempt: Since $f$ us $\mu$-integrable and $\mu(B)< \infty$ for all bounded $B$, notice for any $\varepsilon>0$ there is an $x_0\in \mathbb{R}$ such that $\int_{x \geq x_0} f(x) d \mu(x) < \varepsilon$. Thus for any $\varepsilon$, find such an $x_0$ and find and $N \in \mathbb{N}$ such that for all $n>N$ we have $a_n > x_0$. Then for any $n>N$ we have $\int_{a_n}^{b_n} f d\mu \leq \int_{x \geq x_0} f(x) d\mu(x)< \varepsilon$. Thus $\lim_n \int_{a_n}^{b_n} f d \mu=0$ as desired.

Answer 1

If $f$ is integrable then $fI[f\geq \alpha]\rightarrow 0$ almost everywhere as $\alpha\rightarrow\infty$, then by dominated convergence theorem

$$ \lim_{\alpha\rightarrow\infty}\int fI[f\geq \alpha]d\mu =0 $$ Now $a_n\rightarrow\infty\Rightarrow b_n\rightarrow\infty$,

and consider

$$ \int f(x)I[a_n \leq x \leq b_n] d\mu(x) \\ =\int f(x)I[x\geq a_n]d\mu - \int f(x)I[x\geq b_n]d\mu $$ Then take limit $n\rightarrow\infty$ you will get the desired answer.