This is an exercise from Stein-Sharachi Chap. 3 Exx. 25
$\textbf{Problem Statement}$ Let $E$ be a set of measure zero in $\mathbb{R}^d$. Show that there exists a non-negative integrable $f$ in $\mathbb{R}^d$, so that $$\liminf_{m(B) \to 0} \frac{1}{m(B)} \int_{B} f(y) dy = \infty$$ for each $x \in E$ and $B$ is any ball containing $x$.
$\textbf{My Attempt:}$. From the hint, let $f(x) = \sum_{n=1}^{\infty} \chi_{\mathcal{O}_n} (x)$, where $\mathcal{O}_n \supset E$ and $m ( \mathcal{O}_n) < \frac{1}{2^n}$. We can do this since $E$ has measure zero.
Then $$\int_{\mathbb{R}^d} f(x) dx = \sum_{n=1}^{\infty} \int \chi_{\mathcal{O}_n}(x) = 1$$ and so $f$ is integrable.
I know have left out some details but the above is not where I am having difficulties.
Let $B$ be any ball. From above we have that $E \subset \mathcal{O}_n$. Thus for any $x \in E$, there exists a ball $B_n \subset \mathcal{O}_n$ such that $x \in B_n$.
$$ \int_B f(y) dy = \sum_{n=1}^{\infty} m ( \mathcal{O}_n \cap B) \ge \sum_{n=1}^{\infty} m ( B_n \cap B)$$
Then $$\frac{1}{m(B)} \int_B f(y) dy \ge \sum_{n=1}^{\infty} \frac{m(B_n \cap B)}{m(B)}$$
Since $x \in B_n$ is a point of Lebesgue density I have that $$\lim_{m(B) \to 0} \frac{m(B_n \cap B)}{m(B)} =1$$
This seems useful since then, $$ \sum_{n=1}^{\infty} \lim_{m(B) \to 0} \frac{m(B_n \cap B)}{m(B)} = \infty$$ and I would have my result.
However, how does $\liminf$ ever come in to play? I really do not understand the significance.
And how do I back up moving the limit inside of the summation?
I'm not exactly sure what you're looking for here, but Fatou's lemma might be it! You'd have $$ \sum\limits_{n=1}^{\infty} \liminf_{m(B) \to 0} \frac{m(B_n \cap B)}{m(B)} \leq \liminf_{m(B) \to 0} \sum\limits_{n=1}^{\infty} \frac{m(B_n \cap B)}{m(B)} \leq \liminf_{m(B) \to 0} \frac{1}{m(B)}\int_B f $$