Q: Prove that $\lim\limits_{n\to\infty} \int^{b}_{a}\left(\sum^{n}_{k=1} (f_{k}(x))^{n}\right)^{1/n}dx=\int^{b}_{a}f(x) dx$ I found this question interesting from "Introduction to Analysis by William R. Wade" and decided to share its proof and  also like to learn your approach to proving it.
Supposing that $f_k:[a,b]\to \Bbb{R}$ is continuous for each $k\in \Bbb{N}$, $0\leq f_{k}(x)\leq f_{k+1}(x) ,$ for all $k\in\Bbb{N}$ and $x\in[a,b]$ and $f_k\to f,$ uniformly on $[a,b],$ then 
\begin{align}\lim\limits_{n\to\infty} \int^{b}_{a}\left(\sum^{n}_{k=1} (f_{k}(x))^{n}\right)^{1/n}dx=\int^{b}_{a}f(x) dx\end{align}
PROOF
Let $\epsilon>0$ be given and $x\in [a,b]$. We recall that,
\begin{align} f_n(x)=\left[(f_n(x))^n\right]^{1/n}\leq  \left(\sum^{n}_{k=1} (f_{k}(x))^{n}\right)^{1/n} \leq n^{1/n}f_{n}(x) .\end{align}
For more on the above bound, see Prove that $\left(\sum^{n}_{k=1} (f_{k}(x))^{n}\right)^{1/n}\leq n^{1/n}f_{n}(x)$. Thus,
\begin{align} \left|\left(\sum^{n}_{k=1} (f_{k}(x))^{n}\right)^{1/n}-f(x)\right|\leq \max\{\left|f_n(x)-f(x)\right|, \left|n^{1/n}f_{n}(x)-f(x)\right|\}\end{align}
However, 
\begin{align}\left|n^{1/n}f_{n}(x)-f(x)\right|&=\left|n^{1/n}f_{n}(x)-n^{1/n}f(x)+n^{1/n}f(x)-f(x)\right|\\ &\leq\left|n^{1/n}f_{n}(x)-n^{1/n}f(x)\right|+\left|n^{1/n}f(x)-f(x)\right|\\ &=n^{1/n}\left|f_{n}(x)-f(x)\right|+\left|f(x)\right|\left|1-n^{1/n}\right|\end{align}
Since $f_k:[a,b]\to \Bbb{R}$ is continuous for each $k\in \Bbb{N}$ and $f_k\to f,$ uniformly on $[a,b],$ then $f:[a,b]\to \Bbb{R}$ is continuous and also bounded, i.e., there exists $M>0,$ such that $|f(x)|\leq M,\;\forall\;x\in [a,b]$. Uniform convergence of $f_k\to f,$ on $[a,b]$ implies there exists $N_1(\epsilon)$ such that for all $n\geq N_1(\epsilon),$ $x\in [a,b]$
\begin{align}\left|f_{n}(x)-f(x)\right|<\dfrac{\epsilon}{4}<\epsilon.\end{align}
Also, convergence of $n^{1/n}$ to 1 implies there exists $N_2(\epsilon)$ such that for all $n\geq N_2(\epsilon),$
\begin{align}\left|1-n^{1/n}\right|<\dfrac{\epsilon}{2M}.\end{align}
Hence,
\begin{align}\left|n^{1/n}f_{n}(x)-f(x)\right|&\leq n^{1/n}\left|f_{n}(x)-f(x)\right|+\left|f(x)\right|\left|1-n^{1/n}\right|\\&\leq 2\left(\dfrac{\epsilon}{4}\right)+M \left(\dfrac{\epsilon}{2M}\right)\\&=\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon,\;\;\forall\;n\geq  \max\{N_1(\epsilon),N_2(\epsilon)\}\end{align}
Consequently,
\begin{align} \left|\left(\sum^{n}_{k=1} (f_{k}(x))^{n}\right)^{1/n}-f(x)\right|&\leq \max\{\left|f_n(x)-f(x)\right|, \left|n^{1/n}f_{n}(x)-f(x)\right|\}\\&<\epsilon,\;\;\forall\;n\geq  \max\{N_1(\epsilon),N_2(\epsilon)\}\end{align}
and we are done!

A: My proof: $(\sum_{k=1}^{n} (f_k(x))^{n})^{1/n} \leq f(x) n^{1/n}$ and $n^{1/n} \to 1$ as $n \to \infty$. This gives $\lim \sup$ LHS $\leq$ RHS. Given $\epsilon >0$ choose $N$ such that $f_k(x) >f(x)-\epsilon$ for alll $x$ if $k \geq N$. Noting that $\sum_{k=1}^{n} (f_k(x))^{n})^{1/n} \geq \sum_{k=N}^{n} (f_k(x))^{n})^{1/n} \geq \sum_{k=N}^{n} (f(x)-\epsilon)^{n})^{1/n}$ and $(n-N)^{1/n} \to 1$ as $n \to \infty$ we get $\lim \inf$ LHS $\geq$ RHS $-\epsilon$
