Let $ f_n $ be a sequence of integrable functions in $ [0,1] $ and $ f_n \to f $ uniformly in $ [0,1] $. Let $ b_n $ be an increasing sequence such that $ b_n \to 1 $, when $ n \to \infty $. Prove that $$\lim_{n\to \infty}\int_0^{b_n}f_n(x)\ \text{d}x=\int_0^1 f(x)\ \text{d}x.$$
My try.
Since $f_n\to f$ uniformly, then given $\epsilon>0$, exist $N\in\Bbb N$ such that if $n>N$ we have to $|f_n(x)-f(x)|<\epsilon.$ Now, since $b_n\to 1$, when $n\to \infty$, then exists $N'\in\Bbb N$ such that if $n>N'$ we have to $|b_n-1|<\epsilon$. If $n>\max\{N,N'\}$, then
$$\begin{align*}\left |\int_0^{b_n}f_n(x)\ \text{d}x-\int_0^1 f(x)\ \text{d}x\right |&\leq \left |\int_0^1 f_n(x)\ \text{d}x-\int_0^1 f(x)\ \text{d}x\right |\\ &=\left | \int_0^1 (f_n(x)-f(x))\ \text{d}x \right |\\ &\leq \int_0^1 |f_n(x)-f(x)|\ \text{d}x\\ &<\int_0^1\epsilon \ \text{d}x\\ &=\epsilon.\end{align*}$$
This ends the proof. My proof is right?
Best Answer
First suppose $f \geq 0$, and let $g_n = \chi_{[0, b_n]}f$ where $\chi$ is the indicator function. Then we have $$ |\int_0^{b_n}f_n(x) - \int_0^1f(x)| \leq |\int_0^{b_n} f_n(x) - g_n(x)| + |\int_0^1 g_n(x) - f(x)| $$ The first term on the right tends to $0$ as $n \to \infty$ due to the uniform convergence and the second term tends to $0$ by the monotone convergence theorem.
For the general case, let $f = f^+ - f^-$ where $f^+ = \max(f, 0)$, $f^- = \max (-f, 0)$ and let $f_n = f_n^+ - f_n^-$ in the same way. Note that $$|f^\pm_n(x) - f^\pm(x)| \leq |f_n(x) - f(x)|$$ and so $f_n^\pm \to f^\pm$ uniformly. Hence $$ \int_0^{b_n} f_n^\pm \to \int_0^1 f^\pm. $$ Putting this all togeather, $$ \int_0^{b_n} f = \int_0^{b_n} (f^+_n - f^{-}_n) \to \int_0^1 (f^+ - f^-) = \int_0^1 f. $$