Does there exist $f \in L^1(\mathbb{R})$ where $\lim_{r \to 0} {1\over{r}} \int_{x-r}^{x+r} f(y)\,dy = \infty$?

Question

If $E \subset \mathbb{R}$ has measure $0$, does there exist $f \in L^1(\mathbb{R})$ such that, for every $x \in E$, $$\lim_{r \to 0} {1\over{r}} \int_{x-r}^{x+r} f(y)\,dy = \infty?$$ What if $E$ has positive measure?

Answer 1

We will denote by $|S|$ the Lebesgue measure of any measurable $S\subseteq\mathbb{R}$. Suppose $|E|=0$. Since Lebesgue measure is regular, we can find for any positive integer $k$ an open set $U_k\supseteq E$ such that $|U_k|<2^{-k}$. Now put $f=\sum_{k=1}^\infty\mathbf{1}_{U_k}$ (here $\mathbf{1}_S$ denotes the indicator function of $S$). The monotone convergence theorem tells you that $\int_{\mathbb{R}}f=\sum_{k=1}^\infty\int_{\mathbb{R}}\mathbf{1}_{U_k}<\sum_{k=1}^\infty 2^{-k}<\infty$, so $f\in L^1(\mathbb{R})$.

But now let us fix any $x\in E$ and any integer $N>0$. Since $U_1\cap\cdots\cap U_N$ is an open set containing $x$, we have $(x-r,x+r)\subseteq U_1\cap\cdots\cap U_N$ for $r>0$ sufficiently small, so for such $r$ it holds $$\int_{x-r}^{x+r}f\ge\int_{x-r}^{x+r}\sum_{k=1}^N \mathbf{1}_{U_k} =\int_{x-r}^{x+r}N=2Nr,$$ so (since $N$ was arbitrary) $\lim_{r \to 0}\frac{1}{r}\int_{x-r}^{x+r} f=\infty$.

As for the second question, you need a somewhat more advanced tool from analysis, namely the Hardy-Littlewood maximal function. We have $$\left|\frac{1}{2r}\int_{x-r}^{x+r} f\right|\le\frac{1}{2r}\int_{x-r}^{x+r}|f|\le (Mf)(x),$$ but the maximal inequality by Hardy and Littlewood implies that $(Mf)(x)<\infty$ for a.e. $x$. So the limit that you wrote cannot be $\infty$ for all $x\in E$ (since $|E|>0$). In fact, Lebesgue differentiation theorem (which is a consequence of the maximal inequality) tells you more precisely that we have $$\lim_{r\to 0}\frac{1}{2r}\int_{x-r}^{x+r} f=f(x)$$ for a.e. $x$ (so in particular the limit exists and is finite a.e.).