# Does a function sequence with the integral 1 have a limit with the integral 1?

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?

• Depending on the type of convergence it may or may not be true 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

$$f_n(x)=nxe^{-nx^2}$$

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

But

$$f_n\to 0$$

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 $[a,b]$.