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Theorem $mathbf{A.2.11}$ (Dominated convergence). Let $f_n : X to mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X to mathbb R$ such that $|f_n(x)| leq |g(x)|$ for $mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $mu$-almost every point to some function $f : X to mathbb R$. Then $f$ is integrable and satisfies $$lim_n int f_n , dmu = int f , dmu.$$

I wanted to know if in the hypothesis $|f_n(x)| leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?

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 $xin mathbf R_{ge 0}$, but they are not dominated by an integrable function $g$, and indeed we do not have
$$
lim_{ntoinfty} int f_n = int lim_{ntoinfty}f_n
$$
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 $xge 0$, so it would not be integrable on $mathbf R_{ge 0}$.