Otdfbt

Theorem $mathbfA.2.11$ (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
$$
f_n(x) := frac1n mathbf1_[0,n](x).
$$
Clearly, $f_n in L^1(mathbbR)$ for each $n in mathbbN$. Moreover, $f_n(x) to 0$ as $n to infty$ for each $x in mathbbR$. However,
beginalign*
lim_n to infty int_mathbbR f_n,mathrmdm = lim_n to infty int_0^n frac1n,mathrmdx = 1 neq 0.
endalign*

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 $(X,mathfrakM,mu)$ converging almost everywhere to a measurable function $f$. If $E in mathfrakM$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
$$
lim_n to infty int_E f_n,mathrmdmu = int_E f,mathrmdmu.
$$
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
$$
lim_ntoinfty int f_n = int lim_ntoinftyf_n
$$
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 $mathbf R_ge 0$.