A question on $\liminf$ and $\limsup$ and interchanging limits I have a real function, $f(n,m)$, which is not necessarily bounded nor necessarily non-negative, 
but has point-wise convergence of:
\begin{equation*}
g(m) = \sum\limits_{n=1}^{\infty} f(n,m)
\end{equation*}
and $g(m)$ is finite for all integers $m$.
I am examining the interchange of limits in
\begin{equation*}
\sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} f(n,m) \sim 
\sum\limits_{n=1}^{\infty} \sum\limits_{m=1}^{\infty} f(n,m)
\end{equation*}
and will settle for bounding the summation within a (possibly infinite) range.
If I define:
\begin{equation*}
g_{N}(m) = \inf_{K \ge N} \sum\limits_{k=1}^{K} f(k,m) \le g(m)
\end{equation*}
and 
\begin{equation*}
h_{N}(m) = \sup_{K \ge N} \sum\limits_{k=1}^{K} f(k,m) \ge g(m)
\end{equation*}
Then, can I state the following?
IF $\sum\limits_{m=1}^{\infty} g(m)$ converges, then the limit is within the (possibly infinite) range of:
\begin{equation*}
\begin{aligned}
\liminf_{N \to \infty} \liminf_{M \to \infty} \sum\limits_{m=1}^{M} \sum\limits_{n=1}^{N} f(n,m)
   &\le
\lim_{M \to \infty} \sum\limits_{m=1}^{M} g(m)
\\ &\le
\limsup_{N \to \infty} \limsup_{M \to \infty} \sum\limits_{m=1}^{M} \sum\limits_{N=1}^{N} f(n,m)
\end{aligned}
\end{equation*}
My reasoning is thus:
\begin{equation*}
\begin{aligned}
\liminf_{N \to \infty} \liminf_{M \to \infty} \sum\limits_{m=1}^{M} \sum\limits_{n=1}^{N} f(n,m)
   &\le
\liminf_{M \to \infty} \liminf_{N \to \infty} \sum\limits_{m=1}^{M} \sum\limits_{n=1}^{N} f(n,m)
\\ &\le
\liminf_{M \to \infty} \sum\limits_{m=1}^{M} \sum\limits_{N=1}^{\infty} g_{N}(m)
\\ &\le
\liminf_{M \to \infty} \sum\limits_{m=1}^{M} g(m)
\\ &\le
\lim_{M \to \infty} \sum\limits_{m=1}^{M} g(m)
\\ &\le
\limsup_{M \to \infty} \sum\limits_{m=1}^{M} g(m)
\\ &\le
\limsup_{M \to \infty} \sum\limits_{m=1}^{M} \sum\limits_{N=1}^{\infty} h_{N}(m)
\\ &\le
\limsup_{M \to \infty} \limsup_{N \to \infty} \sum\limits_{m=1}^{M} \sum\limits_{N=1}^{N} f(n,m)
\\ &\le
\limsup_{N \to \infty} \limsup_{M \to \infty} \sum\limits_{m=1}^{M} \sum\limits_{N=1}^{N} f(n,m)
\end{aligned}
\end{equation*}
With the understanding that the 
$\liminf$ or $\limsup$ operations could diverge 
to $\pm \infty$ or converge to different values,
the statement above is similar to Fatou's Lemmas, but with the counting measure and with
a much, much weaker convergence statement.
My question:
Is this weak, weak convergence statement true for all such $f(n,m)$?
If it is not true, can a counter example be provided?
 A: Herea a counterexample:
Let $f(n,m)$ be defined as follows. For fixed $m$, the sequence $f(n,m)$ is
$$f(n,m)=\left\{\begin{array}[l]\   1\ \ \ \text{for}\ \ \ n\leq m\\
-m\ \ \ \text{for }\ \ \ n=m+1\\
0\ \ \ \text{for}\ \ \  n>m+1\end{array}\right.$$
Thus $g(m)=0$ for any $m$ and hence $\sum g(m)=0$.
On the other hand for any $N$,
$$\liminf_{M\to\infty} \sum_{m=1}^M\sum_{n=1}^N f(n,m)=\infty$$ because
$$\sum_{m=1}^M\sum_{n=1}^Nf(n,m)=\sum_{m\leq N}\sum_{n=1}^Nf(n,m)+\sum_{m>N}\sum_{n=1}^Nf(n,m)=0+(M-N)\sum_{n=1}^N 1=(M-N)N$$
therefore
$$\liminf_{N\to\infty}\liminf_{M\to\infty} \sum_{m=1}^M\sum_{n=1}^N f(n,m)=\liminf_{N\to\infty}\infty=\infty$$
and therefore the claim
$$\liminf_{N\to\infty}\liminf_{M\to\infty} \sum_{m=1}^M\sum_{n=1}^N f(n,m)\leq\sum g(m)$$ is false
A: Consider this infinite table:
$$
\begin{pmatrix}
* & 0 & 0 & 0 & 0 & 0 & \dots \\
* & * & 0 & 0 & 0 & 0 & \dots \\
0 & * & * & 0 & 0 & 0 & \dots \\
0 & 0 & * & * & 0 & 0 & \dots \\
0 & 0 & 0 & * & * & 0 & \dots \\
0 & 0 & 0 & 0 & * & * & \dots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}
$$
If you put numbers for the stars you can construct all systems of row and column sums. The sequences $R_n=\sum\limits_{k=1}^\infty f(n,k)$ and $C_n=\sum\limits_{n=1}^\infty f(n,k)$ can be any sequences, independently.
So, without additional assumptions there is no such statement on changing the order of summations.
