This question comes from Tom M. Apostol's Mathematical Analysis I'm not sure which edition, it is the exercise 13-14. The question is the following (my book is in Spanish and I'm translating back to English, so I'm sorry if the translation it's not accurate.)
Let $\lbrace f_n\rbrace$ be a sequence of continuous real functions defined in the interval $[0,1]$ and suppose $f_n\to f$ uniformly on $[0,1]$. Determine if the following is true or not. $$ \lim_{n\to\infty}\int_0^{1-1/n}f_n(x)dx = \int_0^1f(x)dx. $$
I have already solve it, the answer being that it is true, but I want to know if the following approach could work. I'm not looking for an alternate solution.
For each natural number $n$ let $F_n(x) := \int_0^xf_n(t)dt$ and $F:=\int_0^xf(t)dt$. (note that $F_n$ is continuous for every $n$).
Since $f_n\to f$ uniformly, we know that $F_n\to F$ uniformly. (and so $F$ is also continuous).
What we want to prove is the same as $$ \lim_{n\to\infty}F_n(1-1/n)-F_n(0) = F(1)-F(0) $$ We already know that $\lim_{n\to\infty}F_n(0)=F(0)$ so its left to prove that $\lim_{n\to\infty}F_n(1-1/n)=F(1)$.
We have the following \begin{align} F(1) &=\lim_{x\to 1^-}F(x)\\ &=\lim_{x\to 1^-}\lim_{n\to\infty}F_n(x)\\ &=\lim_{n\to\infty}\lim_{x\to 1^-}F_n(x)\\ \end{align} Where the first equality holds because $F$ is continuous, the second one because $F_n\to F$ and the third one because the convergence of $\lbrace{F_n}\rbrace$ is uniform.
Now, since each $F_n$ is continuous we have that
$$
\lim_{x\to1^-}F_n(x) = \lim_{m\to\infty}F_n(1-1/m)
$$
So
$$
F(1) = \lim_{n\to\infty}\lim_{m\to\infty}F_n(1-1/m)
$$
Could we use this to show what we want?
In other words, can we prove that
$$
\lim_{n\to\infty}\lim_{m\to\infty}F_n(1-1/m)=\lim_{n\to\infty}F_n(1-1/n)
$$
Thanks in advanced.
I do not think so. I do not agree with the equality: $$ \lim_{n\to\infty}\lim_{m\to\infty}F_n(1-1/m)=\lim_{n\to\infty}F_n(1-1/n) $$ If you use the Monotone Convergence Theorem, then it holds. However, you need something else before you plug in the equality.