$ \lim_{j\rightarrow \infty}\mu(A_j) \implies \lim_{j\rightarrow \infty}\int_{A_j} u d\mu=0, \forall u\in \mathcal L^1(\mu)$

Question

Let $(X,\mathcal A,\mu)$ be a $\sigma$-finite measure space.

Suppose $(A_j)_{j\in\mathbb N}\subset \mathcal A$ satisfies $\mu(A_j)\rightarrow 0$ as $j\rightarrow \infty$. Show that $$\lim_{j\rightarrow \infty}\int_{A_j} u d\mu=0, \forall u\in \mathcal L^1(\mu)$$

If I'd knew that the sequence $(A_j)_{j\in\mathbb N}\subset \mathcal A$ was increasing then I would have use the fact that $\lim_{j\rightarrow \infty}\mu(A_j)=0$ and the continuity of the measure to see that $\mu( \lim_{j\rightarrow \infty} A_j)=0$ hence $$\lim_{j\rightarrow \infty} A_j=N\in\mathcal N_{\mu}$$

Then using the fact that $u$ is integrable I could have applied Beppo-Levi theorem to $$\lim_{j\rightarrow \infty} \int 1_{A_j}u^{\pm}d\mu$$ leading to

$$\int 1_{N}u^{\pm}d\mu$$ which is the integral of $u$ over a $\mu$-null set and hence equal $0$.

But the problem is that I don't know whether the sequence $(A_j)_{j\in\mathbb N}\subset \mathcal A$ is decreasing, increasing or none.

I was advised to use Vitali's convergent theorem, but I don't really see how :

Let $f_j:=u1_{A_j}$, then it's a sequence of measurable functions,

We see that for any $\epsilon>0$ $$\mu(\{|f_j-0|>\epsilon\}\cap A_j)<\mu(A_j)\rightarrow 0$$ but this does not ensure that the sequence $(f_j)_j$ converges in measure to $0$ basically because to make that assertion we need

$$\mu(\{|f_j-0|>\epsilon\}\cap A)\rightarrow 0, \color{red}{ \forall A\in \mathcal A, \mu(A)<\infty}$$

How can I solve this?

Thanks in advance.

Answer 1

Let $u_N=u\mathbf 1_{\{\left\lvert u\right\rvert\leqslant N\}}$; then $$ \left\lvert\int_{A_j}u \right\rvert\leqslant \left\lvert\int_{A_j}u_N \right\rvert +\left\lvert\int_{A_j}(u-u_N) \right\rvert\leqslant \mu(A_j)N+\int_X\left\lvert u-u_N\right\rvert. $$

Answer 2

It is enough to prove this when $u \geq 0$. [Becasue $u =u^{+}-u^{-}$]. Given $\epsilon >0$ choose $N$ such that $\int_{u >N} u <\epsilon$. This can be justified by Monotone Convergence Theorem or DCT. Then note that $\int_{A_n\cap \{u \leq N\}}u \leq N \mu (A_n) \to 0$. Now split $\int_{A_n} u$ as $\int_{A_n\cap \{u \leq N\}}u+\int_{A_n\cap \{u > N\}}u$ to finish the proof.