Measure Convergence Version of Lebesgue Dominated Convergence Theorem I want to prove that LDCT(Lebesgue Dominated Convergence Theorem) continues to hold if I replace the hypothesis $f_n \to f$ (convergence pointwise) with $f_n\to f$ (convergence in measure):
    $$\int fd\lambda=\lim_{n\to\infty}\int f_nd\lambda.$$
 A: Call $(X,\cal F,\mu)$ the involved measure space. Let $g$ integrable such that $|f_n(x)|\leqslant g(x)$ for almost every $x$.
As $g$ is integrable, denote $X':=\{g\neq 0\}=\bigcup_{n\geqslant 1}\{x,|g(x)|>n^{-1}\}$. Then $X'$ with the induced measure is $\sigma$-finite. Applying this version of dominated convergence theorem, we get that 
$$\int_{X'}fd\mu=\lim_{n\to +\infty}\int_{X'}f_nd\mu.$$
As $X\setminus X'=\{g=0\}\subset \{f=0\}\cup\bigcap_{n\geqslant 1}\{f_n=0\}$, we have $\int_{X'}fd\mu=\int_Xfd\mu$.
So fore each $n$, $\int_{X\setminus X'}fd\mu=\int_{X\setminus X'}f_nd\mu=0$, giving the wanted result. 
A: Here is another proof:
I'll prove that $\displaystyle a_n:=\int_\Omega f_nd\mu\longrightarrow l:=\int_\Omega fd\mu$ as $n$ tends to $\infty$.
By a very useful fact in Analysis, it's enough to prove that for each subsequence of $\{a_n\}$ like $\{a_{n_k}\}$, there is a subsequence of $\{a_{n_k}\}$ like $\{a_{n_{k_l}}\}$ which converges to $l$.
Now, given a subsequence $\{a_{n_k}\}$, we have
$$f_{n_k}\xrightarrow[]{\;\mu\;}f.\tag{I}$$
By this fact, there exists a subsequence of $\{f_{n_k}\}$ like $\{f_{n_{k_l}}\}$ which
$$f_{n_{k_l}}\xrightarrow{\;a.e\;}f.\tag{II}$$
(Note that for $\text{(II)}$ we should have assumed that $\Omega$ is of finite measure, however we can get rid of it as @Leo Lerena pointed in the comments.)
Since $\{f_{n_{k_l}}\}$ is dominated by function $g\in\mathcal{L}^1(\Omega,\mu)$, by the original version of $LDCT$,
$$\int_\Omega f_{n_{k_l}}\xrightarrow{\;a.e\;}\int_\Omega f.\tag{III}$$
That is :
$$a_{n_{k_l}}\xrightarrow{n\rightarrow\infty}l \tag*{$\square$.}$$
Note
The technique of using subsequences, rather than sequences, is one of the most powerful tools of proof !
