$\lim_{n \to \infty} \int_X f_n \, d\mu = \int_X f \, d\mu$ implies $\lim_{n \to \infty} \int_B f_n \, d\mu = \int_B f \, d\mu$ for $B \subseteq X$

Question

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.

Answer 1

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

$$\lim_{n \to \infty} \int_B f_n \, d\mu = \int_B f \, d\mu.$$

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

1. $a_n$ converges
2. 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!)