Exercise 25, Chapter 24 of Spivak's Calculus 3rd Edition Theorem: Let $\{f_n\}$ be sequence of integrable functions on interval $I=[a,b]$ and $f$ be the uniform limit of $\{f_n\}$ on the interval, then prove that $f$ is integrable and $\int_a^b f=\lim_{n\to \infty} \int_a^bf_n$. 
Proof: 
In this case, it is not known before hand that $f$ is integrable (#). However, it can be proven that $f$ is actually integrable.  
It will suffice to show that for every $\epsilon \gt 0$ there exists a partition $P$ of $I$ such that $U(f,P)-L(f,P)\lt \epsilon $, where $U(f,P), L(f,P)$ are upper sum and lower sum respectively as used in Darboux's integrals. 
Since, $f_n$ is (are) integrable, for $\epsilon/3\gt 0$ there exists a partition $P=\{a=y_0,y_1,\cdots, y_n=b\}$ of $I$ such that $U(f_n,P)-L(f_n,P)\lt \epsilon \tag{2}$ and by uniform convergence of $f_n$, we also have that $\exists N $ such that for all $x\in I$ and for all $n\ge N$, we have $|f_n(x)-f(x)|\lt \frac{\epsilon}{3} \tag{3}$ 
$U(f,P)-U(f_n,P)=\sum_{i=1}^{n}(M_i-M_i')\Delta_i=\sum_{i=1}^{n}(M_i-M_i')(y_i-y_{i-1})$, where $M_i=\sup f_n (x)$ on $[y_{i-1}, y_i]$ and $M_i=\sup f(x)$ on $[y_{i-1}, y_i] $. 
Question: How can it be shown that $U(f,P)-U(f_n,P)\lt \epsilon/3$? If it could be shown then similar arguments for lower sum and subsequent use of triangular inequality will prove that $f$ is integrable on $I$.  
(#): If it were known in advance that $f$ is integrable on $I$, then clearly for $\frac{\epsilon}{b-a} \gt 0 \;\;\exists N_\epsilon$ such that for all $x\in I$ and for all $n\ge N_\epsilon$, we have $|f_n(x)-f(x)|\lt \frac{\epsilon}{b-a}  \tag {1}$
Therefore, $|\int_a^bf_n(x)-\int_a^b f(x)|=|\int_a^b(f_n(x)-f(x))|\le \int_a^b|(f_n(x)-f(x))|\le \int_a^b \frac{\epsilon}{b-a} =\epsilon \implies \lim_{n\to \infty}\int_a^bf_n(x)=\int_a^b f(x)$. Proved. 
 A: Note that $f$ is integrable (on that interval) if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that for every partition $P$ of mesh less than $\delta$ you have that $U(f,P) - L(f,P)<\varepsilon$. As you noted, it is enough to split this difference into the following terms :
$$I_n = U(f,P) - U(f_n,P)$$
$$J_n = U(f_n,P) - L(f_n,P)$$
$$K_n = L(f_n,P) - L(f,P)$$
Now, given an $\varepsilon>0$, the first sum will be of the form
$$\sum_{p\in P} (M(p)-M_n(p)) \Delta p$$
where by $M$ and $M_n$ I denote the maximum at each interval $p$ of the partition, and a similar description works for the last difference. For this $\varepsilon>0$, you can choose $n$ large enough so that the terms $M-M_n$ and $m_n-m$ are all dominated by $(b-a)^{-1}\varepsilon/3$ independent of $P$ and $n$. In particular, you can choose some large $N$ and obtain the bound
$$U(f,P) - L(f,P) \leqslant 2\varepsilon /3 + J_N$$
Then, you can take $\delta>0$ so that  $J_N<\varepsilon/3$ (since $f_N$ is integrable), thus achieving the bound you wanted.
A: How can it be shown that $U(f,P) - U(f_n,P) < \epsilon/3$?
We have $f_n \to f$ uniformly on $[a,b]$. For any $\epsilon > 0$ there exists $N$ such that $f(x) - f_n(x) < \frac{\epsilon}{4(b-a)}$ for all $n \geqslant N$ and for all $x \in I_j = [y_{j-1},y_j]$ where $I_j$ is any partition subinterval.
For any $n \geqslant N$ and all $x \in I_j$, we have
$$f(x) < \frac{\epsilon}{4(b-a)} + f_n(x) \leqslant \frac{\epsilon}{4(b-a)} + \sup_{x \in I_j}f_n(x) = \frac{\epsilon}{4(b-a)} +M_n(I_j),$$
and, consequently,
$$M(I_j) = \sup_{x \in I_j}f(x) \leqslant \frac{\epsilon}{4(b-a)} +M_n(I_j)$$
Thus,
$$U(f,P) - U(f_n,P) = \sum_{j=1}^n \left(\, M(I_j)- M_n(I_j)\,\right)\, |I_j|\leqslant \frac{\epsilon}{4(b-a)}\sum_{j=1}^n  |I_j| = \frac{\epsilon}{4} < \frac{\epsilon}{3}$$
