A small doubt about the dominated convergence theorem
This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence $$ f_n(x) := \frac{1}{n} \mathbf{1}_{[0,n]}(x). $$ Clearly, $f_n \in L^1(\mathbb{R})$ for each $n \in \mathbb{N}$. Moreover, $f_n(x) \to 0$ as $n \to \infty$ for each $x \in \mathbb{R}$. However, \begin{align*} \lim_{n \to \infty} \int_{\mathbb{R}} f_n\,\mathrm{d}m = \lim_{n \to \infty} \int_0^n \frac{1}{n}\,\mathrm{d}x = 1 \neq 0. \end{align*}
Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < \infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.
Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,\mathfrak{M},\mu)$ converging almost everywhere to a measurable function $f$. If $E \in \mathfrak{M}$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < \infty$, then $$ \lim_{n \to \infty} \int_E f_n\,\mathrm{d}\mu = \int_E f\,\mathrm{d}\mu. $$ In fact, one has $f_n \to f$ strongly in $L^1(E)$.
In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.
In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = \chi_{[n,n+1]}$ on $\mathbf R_{\ge 0}$ are all integrable, and $f_n(x) \to 0$ for all $x\in \mathbf R_{\ge 0}$, but they are not dominated by an integrable function $g$, and indeed we do not have $$ \lim_{n\to\infty} \int f_n = \int \lim_{n\to\infty}f_n $$ since in this case, the left-hand side is $1$, but the right-hand side is $0$.
To see why there is no dominating function $g$, such a function would have the property that $g(x)\ge 1$ for each $x\ge 0$, so it would not be integrable on $\mathbf R_{\ge 0}$.