Interchanging limit and integration when bounds of integration depend on a parameter This is yet another question on when it is permissible to interchange limits and integrals. I am interested in the situation when bounds of integration depend on some parameter, and then the limit is taken with respect to that parameter.
Suppose one has a function $f(x,t)$ where $t$ is some parameter. Functions $a(t)$ and $b(t)$ give the bounds of integration. Let $a(t) \to a$ and $b(t) \to b$ as $t \to T$. I am interested in sufficient conditions that justify claiming that
$$
\lim_{t \to T} \int_{a(t)}^{b(t)} f(x, t) \, \mathrm{d}x= \int_a^b f(x,T) \, \mathrm{d} x.
$$
Well-known versions of the recipes for interchanging limit and integral (Monotone Convergence and Dominated Convergence Theorems) seem to assume that the region of integration is constant, i.e. $a(t) = a$ and $b(t) = b$ for all $t$. 
 A: I will give a partial answer for the case of $ f $ be continuous on the compact $\mathbb{X}\times\mathbb{T}\subset\mathbb{R}^2$. I hope it can help you. 

Proposition Let $f:\mathbb{X}\times\mathbb{T}\to \mathbb{R}$ is continuous function and  $\mathbb{X}$ and $\mathbb{T}$ compact intervals of $\mathbb{R}$. Let $a:\mathbb{T}\to\mathbb{X}$ and $b:\mathbb{T}\to\mathbb{X}$ continuous function in $T$ such that $\lim_{t\to T}a(t)=a$ and $\lim_{t\to T} b(t)=b$. Then 
  $$
\lim_{t\to T}\int^{b(t)}_{a(t)}f(x,t) \mathrm d x = \int^{b}_{a}f(x,t) \mathrm d x 
$$

Proof By Stone–Weierstrass approximation theorem there is a sequence $\{p_n\}_{n\in\mathbb{N}}$ of polynomials in algebra $\mathcal{P}$  of polynomials (which separates points and conten constants) given by
$$
\mathcal{P}=\left\{p\in C^0(\mathbb{K}\times\mathbb{T}): p(x,t)=\sum_{i=1}^I\sum_{j=1}^{J}c_{ij}t^ix^j,\quad 0<|I|<\infty, 0<|J|<\infty, c_{ij}\in\mathbb{R}\right\}
$$
such that the sequence of polynomials converges uniformly to the function $f$  i.e.
$$
\lim_{n\to\infty}\sup_{(t,x)\in K}\|p_n(t,x)-f(t,x)\|=0. 
$$ 
Supose $a$ and $b$ are continuous functions, $a(t),b(t)\in\mathbb{X}$  and $T\in\mathbb{T}$. Now it is easy to prove that,
$$
\lim_{t \to T}\;
\int_{a(t)}^{b(t)} p(x, t) \, \mathrm{d}x= \int_a^b p(x,T) \, \mathrm{d} x
\quad
\forall p\in\mathcal{P}.
$$
Now use the fact that and properties of uniform convergence in 
\begin{align}
\lim_{t\to T}\int^{b(t)}_{a(t)}f(x,t) \; \mbox{d}x
=
&
\lim_{t\to T}\int^{b(t)}_{a(t)}\lim_{n\to\infty}p_n(x,t) \; \mbox{d}x
\\
=
&
\lim_{t\to T}\lim_{n\to\infty}\int^{b(t)}_{a(t)}p_n(x,t) \; \mbox{d}x
\\
=
&
\lim_{n\to\infty}\lim_{t\to T}\int^{b(t)}_{a(t)}p_n(x,t) \; \mbox{d}x
\\
=
&
\lim_{n\to\infty}\int^{b}_{a}p_n(x,T) \; \mbox{d}x
\\
=
&
\int^{b}_{a}\lim_{n\to\infty}p_n(x,T) \; \mbox{d}x
\\
=
&
\int^{b}_{a}f(x,T) \; \mbox{d}x
\\
\end{align}
