# monotone convergence intergration

I'm trying to prove that

$$\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\int_{[0, \epsilon]}fd\mathbb{P}=0$$ where f is a measurable function that takes values in $[0,\infty]$.

We can't solve it with dominated convergence theorem because we have no information for an intergrable bound.

So the only way, I think is to use monotone convergence theorem.

But in order to apply we need a positive increasing function .And here we got $f_{\epsilon}=\frac{1}{\epsilon}f1_{[0,\epsilon]}$ which clearly doesn't converge in 0 , because $\frac{1}{\epsilon}\rightarrow \infty$.

So is there another way to handle this ??

• Are you sure this is true? What if you test the result with a simple function? Say, integrate over $[0,1]$. Or, perhaps you could specify what $\int_{[0,\epsilon]}$ means. Is it the same as $\int_0^{\epsilon}$ ? – Michael Nov 15 '17 at 1:04
• Yes yes and I found it really odd , I'll check it! – Jonathan1234 Nov 15 '17 at 1:07
• @Michael $\int_{[0,\epsilon]}$ is equivalet to $\int_{f\leq \epsilon}$ because f takes values in $[0,\infty]$ – Jonathan1234 Nov 15 '17 at 1:14
• In that case the answer given below by mich95 should be changed to have $f$ be the identity function over the space $[0,1]$ with uniform probability measure, rather than the constant function. It would be better to give a more clear representation of the problem since nobody knew what you meant. – Michael Nov 15 '17 at 3:38

Take $\mathbb{P}$ to be the uniform probability measure on $[0,1]$, and take $f$ to be the constant function $1$. Then $\frac{1}{\epsilon} \int_{[0,\epsilon]} d \mathbb{P}=1$ for any $\epsilon>0$
• ... and for other $f$, the limit could be anything from $0$ to $\infty$ or might not exist at all. – Robert Israel Nov 15 '17 at 1:08
• but here you mention that $\frac{1}{\epsilon} \int_{[0,\epsilon]} d \mathbb{P}=1$ and not zero – Jonathan1234 Nov 15 '17 at 1:12