TITLE: Convergence in measure imply integrals converge?
QUESTION [5 upvotes]: Suppose $(X, \mathcal{M}, \mu)$ is a measure space such that $\mu(X)<\infty$, and let $(f_{n})_{n\geq 1}$ be a sequence of measurable functions converging in measure to a measurable function $f$. 
Does this imply that $$\int_{X} f d\mu =  \lim_{n\to \infty} \int_{X} f_{n}  d\mu ?$$
All I know is that if $f_{n} \to f$ in measure, then there exists a subsequence $(f_{n_{j}})_{j \geq 1}$ such that $f_{n_{j}} \to f$ $\mu$- almost everywhere which implies the above, but I do not know if this is true for the sequence in general.

REPLY [8 votes]: No, even if the sequence converges pointwise, this is, in general, not satisfied.
Just consider $X=(0,1)$, $\mu$ Lebesgue measure and the sequence
$$f_n(x) := n \cdot 1_{(0,1/n)}(x), \qquad x \in (0,1).$$
Remark: Vitali's convergence theorem implies $$\lim_{n \to \infty} \int f_n \, d\mu = \int f \, d\mu$$ for any sequence $f_n$ (converging to measure in $f$) which is uniformly integrable; see e.g. 
René Schilling: Measures, integrals and martingales, Theorem 16.6
for a proof of (this version of) Vitali's convergence theorem. The assumption on the uniform integrability is in particular satisfied if $|f_n| \leq g$ for some integrable function $g$ (not depending on $n$).