Does a point-wise convergence of measurable functions, imply a convergence in finite measure? This is not true for infinite measures (Pointwise convergence, but not in measure). Is it true for a finite measure?  Namely, let a finite (probability) measure $\mu(\cdot)$. Does a point-wise convergence of $\mu-$measurable functions, $\{f_n(x)\}$, imply convergence in measure $\mu(\cdot)$?
Equivalently, does
$$
\forall x~\lim_{n\to \infty} f_n(x) \to f(x)\,,
$$
imply
$$
\forall \varepsilon ~\lim_{n\to \infty} \mu\big{(}\left\{x~\big{|}~|(f_n(x) - f(x)|>\varepsilon\right\}\big{)} \to 0\,~?
$$
 A: If $f_n \overset{a.e.}{\longrightarrow} f$ and the space has finite measure, by Egoroff's theorem $f_n$ converges to $f$ almost uniformly, and therefore converges in measure to $f$.
A: Yeah this is true. You can use Egoroff, but a direct proof using only DCT is not difficult either.
Let $\epsilon>0$. Define $A_n:=\{x: |f(x)-f_n(x)|>\epsilon\}$. Since $f_n \to f$ pointwise, it follows that $1_{A_n} \to 0$ pointwise. Then notice that all of the functions $1_{A_n}$ are bounded above by the constant function $1$, which is in $ L^1$ (since $\mu$ is finite). By DCT, it follows that $\mu(A_n) = \int 1_{A_n} d\mu \to 0$. But $\mu(A_n) \to 0$ precisely means that $f_n \to f$ in measure.
A: By the reversed Fatou lemma, we have 
$\lim \mu( |f_n-f|\geq \epsilon) = \lim \int 1_{\{ |f_n-f|\geq \epsilon \}}d\mu \leq \int \limsup 1_{\{|f_n-f|\geq \epsilon \}}d\mu = 0.$
$1_{\{\cdot\}}$ is the indicator function (or characteristic function for analysis people). Since indicator is bounded, it is integrable as the measure is finite. The use of Fatou lemma is then justified.  
