Here is a general procedure that the OP may follow:
It is enough to consider functions with real values as all functions involved in the OP (a countable collection of them) are integrable. If OP wants to keep working on the general setting of $\overline{R}$ that is fine, but one may want to use the metric $d(x,y)=|\arctan(y)-\arctan(x)|$ to make $(\overline{R},d)$ a metric space under which $\mathbb{R}$ inherits the classical topology on it. All this is just to conclude that since the $f_n$'s are measurable and $f_n$ converges to $f$ uniformly, and hence pointwise, then $f$ is measurable too: the limit of a sequence of measurable (Borel) functions $f_n$ with values in a complete metric space is also Borel measurable. There may be already several postings on MSE about this.
Now that the measurability of $f$ has been established, the problem now depends on what one means by uniform integrability is defined. Here is present two solutions: one under the more common notion of uniform integrability (due to Hunt) and then another under what the Royden-Fitzpatrick considers as uniform-integrability (which somewhat weaker than Hunt's notion).
About Royden-Fitzpatrick notion of uniform integrability
The OP brought to my attention that Royden's book uses a different notion of uniform integrability. In fact, in my (very) old copy of Royden, H. L., Real Analysis, 3rd edition, Prentice Hall, 1998, there is no mention of uniform integrability. I managed to get hold of the 4th edition (now with Fitzpatrick); compared to the 3rd edition, the 4th one fully reorganized and previous material and included more sections and chapters - sections 4.6 and 5.6 deal with uniform integrability. Unfortunately, the second author used rather weaker notion of uniform integrability. The 5th edition has yet an other reorganization and revamping of material (section 4.6 is no longer, it seems that section 5.1 has all we need).
In any case, from the 4th edition, here is what they (the authors) mean by uniform integrability and tightness:
- In a general space $(X,\mathcal{F},\mu)$ a family $\mathcal{F}\subset L_1$ is uniform integrable if for any $\varepsilon>0$ there is $\delta>0$ such that
$$\sup_{f\in\mathcal{F}}\int_A|f|\,d\mu<\varepsilon\quad\text{whenever}\quad \mu(A)<\delta$$
Remark: In old papers and books, the concept of in (1) is known as equiintegrability. This does not fully coincide with Hunt's notion (below), not even in the case of finite measure spaces.
The other concept given in the textbook is that of tightness (which mimics what measure theorists and probabilistic called tightness of measures on Polish spaces).
- In a general measure space $(X,\mathcal{M},\mu)$, a family $\mathcal{F}$ of measurable functions on is said to be tight if for every $\varepsilon>0$ there is $E\in \mathcal{M}$ such that
$\mu(E)<\infty$ and
$$\sup_{f\in\mathcal{F}}\int_{X\setminus E}|f|\,d\mu <\varepsilon$$
The OP is working within the framework of (1) and (2).
Solution under Royden-Fitzpatrick's framework
Assume then that $f_n$ converges to $f$ a.s.
Tightness implies that for $\varepsilon>0$, there is a set $E$ of finite measure such
$$\sup_n\int_{X\setminus E}|f_n|\,d\mu\leq\frac{\varepsilon}{3}$$
An application of Fatou's lemma yields that $(f_n,f:n\in\mathbb{N})$ is also tight, that is
$$\int_{X\setminus E}|f|\,d\mu\leq\liminf_n\int_{X\setminus E}|f_n|\,d\mu\leq\sup_n\int_{X\setminus E}|f_n|\,d\mu\leq\frac{\varepsilon}{4}$$
By equiintegrability (I use this name to differentiate it from Hunt's uniform integrability), there is $\delta>0$ such that whenever $\mu(A)<\delta$, then $\sup_n\int_A|f_n|\,d\mu<\frac{\varepsilon}{4}$. Another application of Fatou's implies that $(f_n,f:n\in\mathbb{N})$ is also equiibtegrable: $\int_A|f|\,d\mu\leq\frac{\varepsilon}{4}$ whenever $\mu(A)<\delta$.
By Egorov's theorem, there is a set $E_0\subset E$ with $\mu(E\setminus E_0)<\delta$ such that $\|(f_n-f)\mathbb{1}_{E_0}\|_\infty\xrightarrow{n\rightarrow\infty}0$.
Then
\begin{align}
\int_X|f_n-f|\,d\mu&=\int_{E_0}|f_n-f|\,d\mu+\int_{E\setminus E_0}|f_n-f|\,d\mu+\int_{X\setminus E}|f_n-f|\,d\mu\\
&\leq \|(f_n-f)\mathbb{1}_E\|_\infty\mu(E_0)+\frac{\varepsilon}{2}+\frac{\varepsilon}{2}
\end{align}
Therefore
$$\limsup_n\int_X|f_n-f|\,d\mu\leq\varepsilon$$
Letting $\varepsilon\searrow 0$ we conclude that (a) $f$ is integrable and (b) $f_n$ converges to $f$ in $L_1$.
Hunt's notion of uniform integrability
The classical definition of uniform integrability introduced by G. A. Hunt in the late 1960's defines a family $\mathcal{F}\subset L_1(X,\mathcal{M},\mu)$ to be uniform integrability if for any $\varepsilon>0$, there is $g\in L^+_1$ such that
$$\sup_{f\in\mathcal{F}}d(f,[-g,g])<\varepsilon$$
where $[-g,g]$ stands for the collection of all $h\in L_1$ such that $|h|\leq g$ and $d(f,[-g,g])=\inf_{h\in [-g,g]}\|f-h\|_1$.
This is equivalent to saying that $\mathcal{F}$ is uniformly integrable if
$$\inf_{g\in L^+_1}\sup_{f\in \mathcal{F}}\int(|f|-g)_+\,d\mu=0$$
When $(X,\mathcal{M},\mu)$ is a finite measure space, all this is equivalent to either ion the following statements:
- $\mathcal{F}$ is uniformly integrable if
$$\lim_{a\rightarrow\infty}\sup_{f\in\mathcal{F}}\int_{\{|f|>a\}}|f|\,d\mu=0$$
- $\sup_{f\in\mathcal{F}}\int_X|f|\,d\mu<\infty$ and for any $\varepsilon>0$, there is $\delta>0$ such that
$$\sup_{f\in\mathcal{F}}\int_A|f|\,d\mu<\varepsilon\quad\text{whenever}\quad \mu(A)<\delta$$
That is why in most books of measure theory or probability theory either (1) or (2) are presented as definitions for uniform integrability (in the setting of finite measure spaces).
When the space $ (X,\mathcal{M},\mu)$ is not finite, then uniform integrability is equivalent to the following statement:
- $\mathcal{F}$ is uniform integrable if $\sup_{f\in\mathcal{F}}\int_X|f|\,d\mu<\infty$, and there exists a function $h\in L^+_1$ such that for any $\varepsilon>0$, there is $\delta>0$ such that
$$\sup_{f\in\mathcal{F}}\int_A|f|\,d\mu<\varepsilon\quad\text{whenever}\quad \int_A h\,d\mu<\delta$$
If in addition, $(X,\mathcal{M},\mu)$ is $\sigma$-finite, then uniform integrability is equivalent to
- $\mathcal{F}$ is uniform integrable if $\sup_{f\in\mathcal{F}}\int_X|f|\,d\mu<\infty$ and for any $h\in L_1$ such that $h>0$, we have that for any $\varepsilon>0$, there is $\delta>0$ such that
$$\sup_{f\in\mathcal{F}}\int_A|f|\,d\mu<\varepsilon\quad\text{whenever}\quad \int_A h\,d\mu<\delta$$
A detailed discussion of this appears in this posting at MSE. The Wikipedia (English version) page on this feeds in part from the posting at MSE and has the reference to the original paper of Hunt. the German page was ahead of the English version until 2022.
Solution to problem under Hunt's definition of uniform integrability
If the sequence $(f_n:n\in\mathbb{N})\subset L_1$ us uniform integrable (a la Hunt) then $M=\sup_n\|f_n\|_1<\infty$ and the desired conclusion follows by an direct application Fatou's lemma:
$$\int_X|f|\,d\mu=\int_X\lim_n|f_n|\,d\mu\leq\liminf_n\int_X|f_n|\,d\mu\leq M<\infty$$
As with the Royden-Fitzpatrick's framework, we can further show that under Hunt's framework $f_n$ converges to $f$ in $L_1$.
Indeed, the integrability of $f$ and the uniform integrability of $(f_n:\in\mathbb{N})$ implies that $(f_n-f:n\in\mathbb{N})$ is also uniformly integrable. Choose $ g\in L^+_1$ so that $\sup_n \int_{\{|f_n-f|>g\}}|f_n-f|\,d\mu<\frac{\varepsilon}{2}$. As $g_n=|f_n-f|\wedge g\xrightarrow{n\rightarrow\infty}0$ a.s., we have that $\|g_n\|_1\xrightarrow{n\rightarrow\infty}0$ by dominated convergence.
Notice that $\{|f_n-f|>g_n\}=\{|f_n -f|>g\}$; hence,
\begin{align}
|f_n-f|\leq |f_n-f|\mathbb{1}_{\{|f_n-f|>g\}}
+ g_n.\tag{1}\label{one}
\end{align}
Integrating both sides of \eqref{one} and letting $n\nearrow\infty$ gives $\limsup_n\|f_n-f\|_1 \leq \frac{\varepsilon}{2}$. This shows that $\|f_n-f\|_1\rightarrow0$.
