# Non-differentiable in a null set

This is a problem from Stein's real analysis book that I have been working on.

Show that exists a non-negative integrable f in $\mathbb{R}^{d}$ so that $\liminf_{m\left(B\right)\rightarrow0,x\in B}\dfrac{1}{m\left(B\right)}\int_{B}f\left(y\right)dy=\infty\ \ \mbox{for each${x\in E}$}$ where $E$ is measure zero set.

Hint: Find open sets ${\cal O}_{n}\supset E$ with $m\left({\cal O}_{n}\right)<2^{-n}$ and $f\left(x\right):=\sum_{n=1}^{\infty}\chi_{{\cal O}_{n}}\left(x\right)$

I followed the hint and get that by monotone convergence theorem $\liminf\dfrac{1}{m\left(B\right)}\int_{B}\sum_{n=1}^{\infty}\chi_{{\cal O}_{n}}\left(x\right)dx=\liminf\sum_{n=1}^{\infty}m\left({\cal O}_{n}\cap B\right)$ but don't know how to proceed from here.Any help is appreciated.

HINT: if $B$ is small enough you have that for any $M > 0$:
$$\dfrac{1}{m(B)} \int_B \sum \chi_{\mathcal{O}_n} = \dfrac{1}{m(B)} \sum m(\mathcal{O}_n \cap B) > M$$