Limit change of double sequence of functions which diverges Let $(X,\mu)$ be a finite measure space and $\{g_{m,n}(x)\}_{n,m\geq 1}$ be a double sequence of (nonnegative) real valued measurable functions on $X$. Suppose that this sequence satisfies
$$ g_{m,n}(x)\leq g_{m+1, n}(x)\,\,\forall x\in X$$
$$\lim_{n\to\infty} g_{m,n}(x)=c_{m}\,\,\text{for a.e. }x\in X$$
$$\lim_{m\to\infty} c_{m}=\infty$$
$$\lim_{n\to\infty} g_{m,n}(x)=f_{n}(x)\,\,\forall x\in X$$
for some real sequence $\{c_{m}\}_{m\geq 1}$ and sequence of functions $\{f_{n}(x)\}_{n\geq 1}$. Then 
$$\lim_{n\to \infty} f_{n}(x)=\infty$$ for a.e. $x\in X$. 
Is this true? If not, what kind of conditions can be added to make the proposition true? Thanks in advance.
 A: We only need to prove the following lemma : 
Lemma
Let $\{a_{n.m}\}_{n.m\geq 1}$ be a nonnegative sequence with the following properties : 
1) $a_{n,m}$ is increasing w.r.t. $m$, i.e. $a_{n,m}\leq a_{n,m+1}\leq a_{n,m+2}\leq\cdots$. 
2) $\lim_{n\to \infty}a_{n,m}=c_{m}$ for all $n$. 
3) $\lim_{m\to \infty} a_{n,m}=b_{n}$ for all $m$. 
4) $\lim_{m\to\infty} c_{m}=\infty$
Then $\lim_{n\to\infty} b_{n}=\infty$.
proof.
Note that $c_{m}$ is increasing sequence since $a_{n,m}$ is increasing w.r.t $m$. 
Let $L>0$. Since $\lim_{m\to\infty} c_{m}=\infty$, there exists $M_{1}$ s.t. $m\geq M_{1}\Rightarrow c_{m}> 2L$. 
Then for each $m$, there exists $N_{m}$ s.t. $n\geq N_{m}\rightarrow |a_{n,m}-c_{m}|<L\Rightarrow a_{n,m}>c_{m}-L>L$. 
Since $a_{n,m}$ is increasing w.r.t. $m$, $b_{n}\geq a_{n,m}>L$. 
Thus for $N=N_{M_{1}}$, we have $b_{n}>L$ and we get $\lim_{n\to \infty} b_{n}=\infty$. 
Now define 
\begin{align*}
E_{M}=\left\{x\in X\biggr| \lim_{N\to \infty}g_{N,M}(x)=c_{M} \right\} , E=\bigcap_{M=1}^{\infty} E_{M}
\end{align*}
Since $\mu(E_{M})=\mu(X), \mu(E_{M}^{c})=0\Rightarrow \mu(\cup_{M=1}^{\infty} E_{M}^{c})=\mu((\cap E_{M})^{c})=0\Rightarrow \mu(E)=\mu(X)$. 
For $x\in E$, define $$a_{N,M}=a_{N,M}(x)=g_{N, M}(x)$$ then the sequence $\{a_{N,M}\}$ satisfies conditions of lemma 2, so 
$$
\lim_{N\to \infty}\lim_{M\to\infty} a_{N,M}=\lim_{N\to\infty}b_{N}=\infty.
$$
