You essentially proved: Every subsequence of $(f_n)$ has a (sub-)subsequence which converges uniformly to zero (why?).
The first bullet point in the link I provided in a comment asserts that then the sequence $(f_n)$ itself must converge to zero.
Here's why: Suppose $f_{n}$ does not converge uniformly to zero. Then there is an $\varepsilon \gt 0$ and a subsequence $f_{n_j}$ with sup-norm $\|f_{n_j}\|_{\infty} \geq \varepsilon$ for all $j$. This subsequence has again a convergent subsequence by Arzelà-Ascoli. Your argument shows that its limit must be zero, hence the subsequence must have uniform norm $\lt \varepsilon$ eventually, contradiction.
Here's the abstract thing:
Let $x_n$ be a sequence in a metric space $(X,d)$. Suppose that there is $x \in X$ such that every subsequence $(x_{n_j})$ has a subsubsequence $(x_{n_{j_k}})$ converging to $x$. Then the sequence itself converges to $x$.
Edit: As leo pointed out in a comment below the converse is also true: a convergent sequence obviously has the property that every subsequence has a convergent subsubsequence.
The proof is trivial but unavoidably uses ugly notation: Suppose $x_n$ does not converge to $x$. Then there is $\varepsilon \gt 0$ and a subsequence $(x_{n_j})$ such that $d(x,x_{n_j}) \geq \varepsilon$ for all $j$. By assumption there is a subsubsequence $x_{n_{j_k}}$ converging to $x$. But this means that $d(x,x_{n_{j_k}}) \lt \varepsilon$ for $k$ large enough. Impossible!
The way this is usually applied in "concrete situations" is to show
- If a subsequence converges then it must converge to a specific $x$. This involves an analysis of the specific situation—this is usually the harder part and that's what you did.
- Appeal to compactness to find a convergent subsubsequence of every subsequence—that's the trivial part I contributed.
For a counterexample, take $f(x) = \cos x$, $a = 0$, $b = \infty$, and $\beta_n = n\pi$.
We have $\beta_n \to \infty$, and
$$\lim_{n \to \infty}\int_a^{\beta_n}f(x) \, dx = \lim_{n \to \infty}\int_0^{n\pi} \cos x\,dx = \lim_{n \to \infty}(\sin n\pi - \sin 0) = 0$$
However, the improper integral $\displaystyle\int_0^\infty \cos x \, dx$ does not converge, since $\underset{\beta \to \infty}\lim \sin \beta$ does not exist.
If $f \geqslant 0$, then $F(\beta)$ is nonnegative and nondecreasing. Suppose there exists a sequence $\beta_n$ where $\beta_n \to \infty$ and
$$\lim_{n \to \infty} F(\beta_n) = \lim_{n \to \infty}\int_a^{\beta_n} f(x) \, dx = I$$
For any $\epsilon > 0$, there exists $N$ such that $|F(\beta_n) - I| < \epsilon$ for all $n \geqslant N$. If $\beta > \beta_N$, then since $\beta_n \to \infty$, there exists $m > N$ such that $F(\beta_N) \leqslant F(\beta) \leqslant F(\beta_m),$ and
$$-\epsilon < F(\beta_N) - I < F(\beta)-I < F(\beta_m)- I < \epsilon$$
Thus,
$$\lim_{\beta \to \infty} F(\beta) = \lim_{\beta \to \infty}\int_a^{\beta} f(x) \, dx = I$$
Best Answer
$g(x)$ has a bounded derivative on $[0,1]$ so by the MVT $g(x)$ is Lipschitz-continuous on $[0,1.$ So $g(f_n)$ converges uniformly to $g(f)$ on $[0,1$. And each $g(f_n)$ is integrable so $g(f)$ is integrable. Hence $$\lim_{n\to \infty}\int_0^1g(f_n(x))dx=\int_0^1g(f(x)dx=\int_0^1\lim_{n\to \infty}g(f_n(x))dx.$$
You cannot prove this without an interchange of the order of the limit and the integral because what you are trying to prove is exactly that the order $can$ be interchanged.