2
$\begingroup$

Let $(f_n)_{n\in\mathbb{N}}$ ne a sequence of continuous functions on $[0,1]\rightarrow\mathbb{R}$ that converges uniformly to $f:[0,1]\rightarrow\mathbb{R}$. Let $(b_n)_{n\in\mathbb{R}}$ be an increasing sequence of real numbers in $(0,1)$ that converges to 1. Prove that : $$\underset{n\rightarrow \infty}{\lim}\int_{0}^{b_{n}}f_n(x)dx=\int_0^1f(x)dx$$

As far as I know if we have a sequence of functions converging uniformly to $f$ AND if they are uniformly bounded we can have the following: $$\underset{n\rightarrow \infty}{\lim}\int_{0}^{1}f_n(x)dx=\int_0^1f(x)dx$$.
But here it is not so. Even the uniform boundedness is also not mentioned.

$\endgroup$
  • 1
    $\begingroup$ Uniform convergence is sufficient for the interchange of limit with the integral. $\endgroup$ – Math1000 Jan 5 at 22:12
  • $\begingroup$ Thank you for correcting $\endgroup$ – DD90 Jan 5 at 22:36
4
$\begingroup$

Hint:

For any $\epsilon > 0$, we have for all sufficiently large $n$ $$\left|\int_0^{b_n} f_n - \int_0^1 f \right| \leqslant \int_0^{b_n}|f_n - f| + \int_{b_n}^1 |f| \leqslant \epsilon b_n + \sup_{x \in [0,1]} |f(x)| (1 - b_n)$$

$\endgroup$
  • $\begingroup$ Thanks. It works. But I need a small clarification: is it always true that $|\int f dx|\leq\int|f|dx$? If so can you please point out a website or a book which state this result? Thanks again! $\endgroup$ – DD90 Jan 5 at 22:35
  • 1
    $\begingroup$ So $-|f(x)| \leqslant f(x) \leqslant |f(x)|$. One side gives us $\int_0^1f \leqslant \int_0^1|f|$. The other side gives us $-\int_0^1|f| \leqslant \int_0^1 f$. Thus $|\int_0^1 f| \leqslant \int_0^1 |f|$. $\endgroup$ – RRL Jan 5 at 22:38
  • $\begingroup$ Ah! So it is the proof. Thank you! $\endgroup$ – DD90 Jan 5 at 22:41
  • 1
    $\begingroup$ @DD90: You are welcome. $\endgroup$ – RRL Jan 5 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.