From https://en.wikipedia.org/wiki/Dominated_convergence_theorem
"In measure theory, Lebesgue's dominated convergence theorem provides sufficient conditions under which almost everywhere convergence of a sequence of functions implies convergence in the $L^1$ norm. Its power and utility are two fo the primary theoretical advantage of Lesbesgue integration over Riemann integration.
"It is widely used in probability theory, since it gives a sufficient condition for the convergence of expected values of random variables."
Lebesgue's Dominated Convergence Theorem Let $\{ f_n \}$ be a sequence of real-valued measurable functions on a measurement space $(S, \Sigma, \mu )$. Suppose that the sequence converges pointwise to a function $f$ and is dominated by some integrable function $g$ in the sense that:
$$ |f_n(x)| \le g(x) $$for all numbers $n$ in the index set of the sequence and all points $x \in S$. Then $f$ is integrable and
$$ \lim_{n \rightarrow \infty} \int_S \left| f_n - f \right|\,d\mu = 0 $$which also implies:
$$ \lim_{n \rightarrow \infty} \int_S f_n \,d\mu = \int_S f\,d\mu $$Remark 1 "The statement 'g is integrable' is meant in the sense of Lebesgue, ie:
$$ \int_S |g|\,d\mu < \infty $$Remark 2 "The convergence of the sequence and domination by $g$ can be relaxed to hold only $\mu$-almost everywhere provided the measure space $(S, \Sigma, \mu)$ is complete or $f$ is chosen as a measurable function which agrees $\mu$-almost everywhere with the $\mu$-almost everywhere existing pointwise limit. (These precautions are necessary, because otherwise there might exist a non-measurable subset of a $\mu$-null set $N \in \Sigma$, hence $f$ might not be measurable).
Remark 3 "If $\mu(S) < \infty$, the condition that there is a dominating integrable function $g$ can be relaxed to uniform integrabiity of the sequence $\{f_n \}$, see Vitali convergence theorem."
Theorem 4.3.1 (Lebesgue dominated convergence theorem) "Suppose $f_n: \mathbb{R} \rightarrow [-\infty, \infty]$ are (Lebesgue) measurable functions such that the pointwise limit $f(x) = \lim_{n \rightarrow \infty} f_n(x)$ exists. Assume there is an integrable $g: \mathbb{R} \rightarrow [0, \infty]$ with $|f_n(x) | \le g(x)$ for each $x \in \mathbb{R}$. Then $f$ is integrable as is $f_n$ for each $n$, and:
$$ \lim_{n \rightarrow \infty} \int_\mathbb{R} f_n\, d\mu = \int_\mathbb{R} \lim_{n \rightarrow \infty} f_n\,d\mu = \int_\mathbb{R} f\,d\mu $$Theorem 4.4.1 (Continuity of integrals) "Assume $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is such that $x \mapsto f^{[t]}(x) = f(x,t)$ is measurable for each $t \in \mathbb{R}$ and $t \mapsto f(x,t)$ is continuous for each $x \in \mathbb{R}$. Assume also that there is an integrable $g:\mathbb{R} \rightarrow \mathbb{R}$ with $\left|f(x,t)\right| \le g(x)$ for each $x,t\in\mathbb{R}$. Then the function $f^{[t]}$ is integrable for each $t$ and the function $F: \mathbb{R} \rightarrow \mathbb{R}$ defined by:
$$ F(t) = \int_\mathbb{R} f^{[t]}\,d\mu = \int_\mathbb{R} f(x,t) \,d\mu(x) $$... is continuous.
Theorem 4.4.3 (Differentiating under the integral sign) "Assume $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is such that $x \mapsto f^{[t]}(x) = f(x,t)$ is measurable for each $t \in \mathbb{R}$, that $f^{[t_0]}(x) = f(x,t_0)$ is integrable for some $t_0 \in \mathbb{R}$ and $\frac{\partial f(x,t)}{\partial t}$ exists for each $(x,t)$. Assume also that there is an integrable $g : \mathbb{R} \rightarrow \mathbb{R}$ with $\left| \left. \frac{\partial f}{\partial t} \right|_{x,t}\right| \le g(x)$ for each $x,t\in \mathbb{R}$. Then the function $x \mapsto f(x,t)$ is integrable for each $t$ and the function $F: \mathbb{R} \rightarrow \mathbb{R}$ defined by:
$$ F(t) = \int_\mathbb{R} f_t\,d\mu = \int_\mathbb{R} f(x,t)\,d\mu(x) $$is differentiable with derivative:
$$ F'(t) = \frac{d}{dt}\int_\mathbb{R} f(x,t)\,d\mu(x) = \int_\mathbb{R} \frac{\partial}{\partial t} f(x,t)\,d\mu(x) $$