Let $f_{t}(x)$ be a compactly supported positive function for all $t\geq 0$. If we know that $\lim_{t\rightarrow \infty}f_{t}(x)$ does not exist for all $x$ (i.e. point-wise), can it be shown that $\lim_{t\rightarrow \infty}\int f_{t}(x)dx$ does not exist?
If the answer is no, what would be the necessary and sufficient conditions on $f_{t}(x)$ so that the answer is affirmative?
It seems that the key is to prove that for every $\varepsilon > 0$ there exists a $t^{'}$ such that for all $t^{'}$ such that for all $x$ in the support of $f_{t^{'}}(x)$ we have $f_{t^{'}}(x)>\frac{\varepsilon}{\mathbf{supp}\big(f_{t^{'}}(x)\big)}$. With the latter, it can then be shown that
$$ \int f_{t^{'}}(x)dx>\varepsilon $$
Thank you very much for your time.
There are counterexamples:
Consider $f_t(x)=t+1$ if $0\leq x\leq1/(t+1)$ and $0$ otherwise, then $f_t\geq0$ is compactly supported and $\lim_{t\to\infty}f_t(x)=\infty$ for all $x\in$ supp$(f_t)$, but $\int f_t(x)=1$ for all $t\geq0$.
There exist many sufficient conditions, here is an easy one: if $f_t\geq0$ for all $t$ and there exists a positive-measure set $S$ such that $\lim_{t\to\infty}f_t(x)=\infty$ for all $x\in S$, then $$\int f_t(x)dx\geq|S|\inf_{x\in S}f_t(x)\to\infty\ \ \text{as}\ \ t\to\infty.$$
Note that this is clearly not necessary. In fact, I feel it's quite hopeless to find necessary and sufficient conditions since the setting is too general.