Epsilon/lambda of simple function

Question

In my homework I have the following function:

$u(x)=\sum_{n=1}^\infty \frac{1}{n^2(n+1)}1_{[-n,n]}$

I want to show that $\lambda(\{u\geq \epsilon\}) \leq \frac{2}{\epsilon}$

I have tried doing the calculations in Excel and it is indeed true.

But I am out of thoughts on how to show this. Can anyone provide some help/hint? Thank you

Answer 1

$\int u(x) dx=\sum \frac {2n} {n^{2}(n+1)} =2\sum (\frac 1n -\frac 1 {n+1})$. This is a telescopic sum and you get $\int u(x) dx=2$. Hence $\lambda(u \geq \epsilon) \leq \frac 1 {\epsilon} \int u(x)dx=\frac 2 {\epsilon}$.

I have used the following standard argument: $\int u(x)dx \geq \int_{u \geq \epsilon} u(x)dx \geq \int_{u \geq \epsilon} \epsilon dx=\epsilon\lambda (u \geq \epsilon)$