Let $(x_n)$ be any function sequence such that

$$ \int_0^1x_n(t)dt=1 \qquad \forall n $$

$$ \lim_{n\to\infty}x_n = x $$

I'm trying to prove that the limit $x$ also has the property $\int_0^1x(t)dt=1$. I don't think I could construct a "bounding" function to use the dominated convergence theorem. Could I have a hint?

  • 3
    $\begingroup$ Depending on the type of convergence it may or may not be true $\endgroup$
    – Mercy King
    Nov 7 '12 at 21:03

You can't prove it because it's not true. Take $x_n=n\chi_{(0,1/n]}$ and $x(t)=0$.


Assuming that you're talking about a pointwise limit, this doesn't work. Consider $$x_n(t)=\begin{cases}-2n^2t+2n & 0<t\leq \frac1n\\0 & \text{otherwise}.\end{cases}$$ These converge pointwise to the zero function.

Now, if we happen to know that the functions $x_n$ converge uniformly, then we can let the limit "pass through" the integral, and get $$1=\lim_{n\to\infty} 1=\lim_{n\to\infty}\int_0^1x_n(t)\,dt=\int_0^1\left(\lim_{n\to\infty}x_n(t)\right)\,dt=\int_0^1 x(t)\,dt.$$


This is not necessarily true. Consider the function


Note that $$\int_0^1 {{f_n}} (x)dx = \int_0^1 {nx{e^{ - n{x^2}}}dx} = \left( { - \frac{1}{2}} \right)\left. {{e^{ - n{x^2}}}} \right|_0^1 = - \frac{1}{2}{e^{ - n}} + \frac{1}{2}$$

Then $$\lim \int_0^1 {{f_n}} (x)dx=\frac 1 2$$


$$f_n\to 0$$

so $$\int_0^1 f(x) dx =0$$

A sufficient condition for this to be true is that the $f_n$ are all integrable and $f_n$ converges uniformly to $f$ over $[a,b]$.


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.