Let $(f_k)$ be a uniformly convergent sequence of bounded functions on $[0,1]$. Suppose that the sequence $(f_k)$ converges uniformly on $[0,1]$ to a function $f$. Determine if

$\lim_{n \to +\infty}\int_0^{1-\frac{1}{n}} f_n(x)\,dx=\int_0^1 f(x)\,dx.$

So far i have proved that $f$ is continuous on the interval, however am confused

I can prove that $\lim_{n \to +\infty}\int_0^{1-\frac{1}{n}} f_n(x)\,dx=\int_0^{1-\frac{1}{n}} f(x)\,dx.$

Does the fact that the interval lies in $[0,1]$ make a difference as $\frac{1}{n}$ of $0,1$ is either $0$ or undefined, and do i use this with the FTC?

  • $\begingroup$ Express the integral as a Riemann sum. $\endgroup$
    – lemon
    Apr 25, 2014 at 16:33
  • $\begingroup$ What you've written on the fifth line makes no sense (having a limit on the left and the integral to $1-1/n$ on the right). You can't take limit in only one of the two places the $n$ appears. $\endgroup$ Apr 25, 2014 at 16:41

1 Answer 1


Assuming all the functions in question are integrable, it should follow pretty easily from the triangle inequality.

Hint: First, show that the functions must be uniformly bounded (i.e., there is $M$ so that $|f_n(x)|\le M$ for all $n$ and all $x$). Now show the following: $$\left|\int_0^{1-1/n} f_n(x)dx - \int_0^1 f(x)dx\right| \le \int_0^1 |f_n(x)-f(x)|\,dx + \int_{1-1/n}^1 |f_n(x)|\,dx\,.$$

  • $\begingroup$ Yes, but how would i get rid of that far right integral ? $$\int_{1-1/n}^1 |f_n(x)|\,dx\,.$$ $\endgroup$
    – user112365
    Apr 27, 2014 at 16:42
  • 2
    $\begingroup$ With the "First" sentence in my hint :) $\endgroup$ Apr 27, 2014 at 16:46
  • $\begingroup$ Your the man. Thank you. $\endgroup$
    – user112365
    Apr 27, 2014 at 17:18
  • $\begingroup$ You're very welcome. $\endgroup$ Apr 27, 2014 at 17:22

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