Let $f:[0,1]\times[0,1]\to\mathbb{R}\cup\{\pm\infty\}$ be a function such that, for a given point $\hat{x}\in(0,1)$, $f$ is continuous in $[0,\hat{x})\times[0,\hat{x})$ and $f(\hat{x},\hat{x})=\infty$. Also let $x_{\varepsilon}\in[0,\hat{x})$ be a sequence such that $x_{\varepsilon}\to\hat{x}$ as $\varepsilon\to0$. We define $$ F(\varepsilon):=\int_{0}^{x_{\varepsilon}}f(x_{\varepsilon},y)dy\quad\text{and}\quad F^{\delta}(\varepsilon):=\int_{0}^{x_{\varepsilon}-\delta}f(x_{\varepsilon},y)dy $$ for a sufficiently small $\delta>0$ which is independent on $\varepsilon>0$.
I'm considering the following problem:
Does the equality $$ \lim_{\varepsilon\to0}F(\varepsilon)=\lim_{\delta\to0}\lim_{\varepsilon\to0}F^{\delta}(\varepsilon)=\int_{0}^{\hat{x}}f(\hat{x},y)dy $$ hold?
Since the dominated convergence theorem (DCT in short) does not work directly for $F(\varepsilon)$, I consider the function $F^{\delta}(\varepsilon)$ so that DCT is valid. But I don't know whether or not the equality is true.
I'm gald if you give some hints or proof to justify.
Let $x_\epsilon$=$\hat x-\epsilon.$ For every $\epsilon>0$ consider a function $\phi_\epsilon$ on $[0,1]$ that has a sharp negative spike with area $-\frac1\epsilon$ concentrated on $[0,\epsilon],$ a sharp positive spike with area $+\frac1\epsilon$ concentrated on $[\hat x-2\epsilon,\hat x-\epsilon]$ a sharp rise to infinity with area $+\epsilon$ near $\hat x,$ and 0 on the interval $[\epsilon,\hat x-2\epsilon].$ It is not difficult to construct such a family with the additional condition that
$$f(x,y):=\phi_x(y)$$
is continuous or even infinitely differentiable.
The sharp rise at the end ensures that the limit for $y$ separately is infinity. But $F(\epsilon)=0$ and $F^\delta(\epsilon)=-\frac1\epsilon$ for all sufficiently small $\epsilon>0.$