$\lim_{n \to \infty} \int_X f_n \, d\mu = \int_X f \, d\mu$ implies $\lim_{n \to \infty} \int_B f_n \, d\mu = \int_B f \, d\mu$ for $B \subseteq X$ I'm having trouble with the following problem.
Let $(X, \mathcal{M},\mu)$ be a measure space, where $X = [a,b] \subset \mathbb{R}$ is a closed and bounded interval and $\mu$ is the Lebesgue measure. Let $f_{n}$ be a sequence of non-negative functions in $L^{1}(X,\mathcal{M},\mu)$ $\textit{converging in measure}$ to a function $f \in L^{1}(X,\mathcal{M},\mu)$. Given that the following holds,
$\lim\limits_{n\rightarrow\infty}\int\limits_{X}f_{n}d\mu = \int\limits_{X}fd\mu$
show that for all $B \subset X$, 
$\lim\limits_{n\rightarrow\infty}\int\limits_{B}f_{n}d\mu = \int\limits_{B}fd\mu$
where $B$ belongs to the Borel $\sigma$-algebra. 
I was given a hint where convergence in measure in X implies convergence in measure in B, but I'm not sure where to proceed from here. 
 A: Let $(f_{n(k)})_{k \in \mathbb{N}}$ be an arbitrary subsequence of $(f_n)_{n \in \mathbb{N}}$. Since $f_n$ converges in measure to $f$, we can choose a subsequence of $(f_{n(k)})_{k \in \mathbb{N}}$ which converges almost everyhwere to $f$; for simplicity of notation we denote this subsequence by $(g_n)_{n \in \mathbb{N}}$.
It follows from Fatou's lemma that
$$\int_B f \, d\mu = \int_B \lim_{n \to \infty} g_n \, d\mu \leq \liminf_{n \to \infty} \int_B g_n \, d\mu. \tag{1}$$
Similarly,
$$\int_{X \backslash B} f \, d\mu \leq \liminf_{n \to \infty} \int_{X \backslash B} g_n \, d\mu.$$
Since $\int_X f \, d\mu = \lim_n \int_X g_n \, d\mu$, this yields
$$\begin{align*} \int_B f \, d\mu &= \int_X f \, d\mu - \int_{X \backslash B} f \, d\mu \\ &\geq \lim_{n \to \infty} \int_X g_n \, d\mu - \liminf_{n \to \infty} \int_{X \backslash B} g_n \, d\mu \\ &= \limsup_{n \to \infty} \int_B g_n \, d\mu. \tag{2} \end{align*}$$
Combining $(1)$ and $(2)$ gives
$$\int f \, d\mu \leq \liminf_{n \to \infty} \int_B g_n \, d\mu \leq \limsup_{n \to \infty} \int_B g_n \, d\mu \leq \int f \, d\mu.$$
Hence,
$$\lim_{n \to \infty} \int_B g_n \, d\mu = \int_B f \, d\mu.$$
Since the limit does not depend on the subsequence (see the lemma below), this already proves
$$\lim_{n \to \infty} \int_B f_n \, d\mu = \int_B f \, d\mu.$$

Lemma (Subsequence principle): Let $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}$. Then the following statements are equivalent:


*

*$a_n$ converges

*For any subsequence of $(a_n)_{n \in \mathbb{N}}$ there exists a subsequence which converges to $a \in \mathbb{R}$. ($a$ does not depend on the subsequence!)

