Lebesgue integral on interval [a,b] Suppose $f$ and $g$ are non negative real functions defined on interval $[a,b]$.  $f$,$g$ $\in L^1([a,b])$
If there is a sequence of decreasing measureable sets $...\subset A_2\subset A_{1}$ ,such that $A_n \subset [a,b]$ for all n. 
On which $\int_{A_n}f d\lambda = \int_{A_n}g d\lambda $ 
Prove for $A = \bigcap_i A_i$,we have $\int_A f d\lambda = \int_A g d\lambda$.
Sorry for the elementary of the question, my attempt is using DCT,since $f\mathbb{1}_{A_n} \le f$ and $f \in L^1$.since $f\mathbb{1}_{A_n} \to f\mathbb{1}_A $ we have  $\int_A f d\lambda = \int_A g d\lambda$ right?
 A: the measure $$\lim_{n \to \infty}\mu(A_n \bigcap A^c)=0$$
$$\lim_{n \to \infty}\int_A (f-g)d\lambda= \lim_{n \to \infty}(\int_{A_n} (f-g)d\lambda - \int_{A_n\bigcap A^c} (f-g)d\lambda)=0$$
A: Your answer is essentially correct. I would only suggest also mentioning that the same argument tells you $\int_A g \mathrm{d}\lambda = \lim_{n \to \infty} \int_{A_n} g \mathrm{d} \lambda = \lim_{n \to \infty} \int_{A_n} f \mathrm{d} \lambda$, but that’s more my comments as a grader. I think you have the right idea
Here’s another, slightly more elementary but less snappy proof. It’s a useful fact that given a function $h \in L^1$, for every $\epsilon > 0$ exists $\delta > 0$ such that if $S$ is measurable and $\lambda(S) < \delta$, then $\int_S |h| \mathrm{d} \lambda < \epsilon$. To prove this, consider some nonnegative simple function $\phi$ for which $0 \leq \phi \leq |h|$ and $\int \phi \mathrm{d} \lambda \geq \int |h| \mathrm{d} \lambda - \epsilon/2$. If $A = \sup \phi < \infty$, then set $0 < \delta < \epsilon/(2A)$.
To use this here, fix $\epsilon > 0$, then choose $\delta_1, \delta_2 > 0$ such that
\begin{align*}
\lambda(S) & < \delta_1 & \Rightarrow \int_S |f| \mathrm{d} \lambda & < \epsilon / 2, \\
\lambda(S) & < \delta_2 & \Rightarrow \int_S |g| \mathrm{d} \lambda & < \epsilon / 2.
\end{align*}
Set $\delta = \min \{\delta_1, \delta_2 \}$. You can prove that $\lambda(A) = \lim_{n \to \infty} \lambda(A_n)$, so there exists $N \in \mathbb{N}$ such that $n \geq N \Rightarrow \lambda(A_n \setminus A) < \delta$. Thus if $n \geq N$, we can estimate
\begin{align*}
\left| \int_A f \mathrm{d} \lambda - \int_A g \mathrm{d} \lambda \right| & \leq \left| \int_A f \mathrm{d} \lambda - \int_{A_n} f \mathrm{d} \lambda \right| + \left| \int_{A_n} f \mathrm{d} \lambda - \int_{A_n} g \mathrm{d} \lambda \right| + \left| \int_{A_n} g \mathrm{d} \lambda - \int_A g \mathrm{d} \lambda \right| \\
& = \left| \int_{A_n \setminus A} f \mathrm{d} \lambda \right| + 0 + \left| \int_{A_n \setminus A} g \mathrm{d} \lambda \right| \\
& \leq \int_{A_n \setminus A} |f| \mathrm{d} \lambda + \int_{A_n \setminus A} |g| \mathrm{d} \lambda \\
& < \epsilon .
\end{align*}
EDIT: Though your DCT argument would also work if we switched $[a, b]$ with any arbitrary measure space, mine breaks down if $\lambda(A_n) = \infty$ for all $n$. This isn’t a problem on the compact interval $[a, b]$, but could be on, say, the real line.
