Hausdorff measure of a subset of $\mathbb R^3$ Let $f \in L^1_{\text{loc}}(\mathbb R^3)$. We define $A \subset \mathbb R^3$ as $$ A := \left\{ x \in \mathbb R^3 \, : \, \limsup_{r \to 0} \frac 1 r \int_{\mathbb B(x,r)} \vert f(y) \vert \, \mathrm dy  > 0\right\} \; .$$
Now the claim is, that $\mathcal H^1(A) = 0$. 
Unfortunately, I have no idea how to show that. Has someone a hint?
 A: Hints:


*

*Use the Lebesgue differentiation theorem to show $\mathcal{L}^3(A) = 0$ 
(here $\mathcal{L}^3$ stands for Lebesgue measure on $\mathbb{R}^3$).

*Fix $\sigma > 0$ and define the subset $A_\sigma \subset A$ where $\limsup_{r \to 0}\frac{1}{r}\int_{B(x,r)}|f(y)|d y > \sigma$. Note that $\mathcal{L}^3(A_\sigma) = 0$.

*Recall that if $|f|$ is (locally) integrable in $\mathbb{R}^3$, then the integral of $|f|$ over a set with small $\mathcal{L}^3$ measure is small (make this rigorous).

*Using 2, there is an open set $U$ containing $A_\sigma$ with small $\mathcal{L}^3$ measure. By 3, the integral of $|f|$ over $U$ is small, say $\int_{U}|f| < \epsilon$ for some $\epsilon > 0$.

*Fix $\delta > 0$. Define a covering of $A_\sigma$ using closed balls $B(x,r) \subset U$ where $x \in A_\sigma$ and $0 < r < \delta/10$. (Part of your definition should include some lower bound for $\frac{1}{r}\int_{B(x,r)}|f(y)|d y$. Hint: it involves $\sigma$). 
Idea: We want to these to be an admissible collection in the definition of $\mathcal{H}^1_\delta(A_\sigma)$, where
\begin{equation}
\mathcal{H}^1_\delta(A_\sigma) := \inf\{\sum_{i = 1}^{\infty}\frac{\text{diam} C_j}{2} \,: A_\sigma \subset \bigcup_{i = 1}^\infty C_j, \text{diam} C_j \leq \delta\}.
\end{equation}

*Use Vitali's covering theorem and step $5$ to find a countable covering - this covering is then admissible in the definition of $\mathcal{H}^1_{\delta}(A_\sigma)$.

*Using step $4, 5$, you should find an estimate which is something like $\mathcal{H}^1_{\delta}(A_\sigma) \leq C \frac{\epsilon}{\sigma}$. Send $\delta \to 0$ and then $\epsilon \to 0$ to find $\mathcal{H}^1(A_\sigma) = 0$.

*Recall the definition of $A_\sigma$, and use the fact that $\mathcal{H}^1(A_\sigma) = 0$ by step 7, to show $\mathcal{H}^1(A) = 0$.
Bonus exercise: The above guide works in any dimension $n$, and any $\mathcal{H}^s$ where $0 < s < n$.
A: Thank you Lost in a Maze for your answer. This helped a lot. I have two points, where I'm not 100% sure. 
At point (1) we have by Lebesgue-Besicovitch
$$\vert f(x) \vert = \limsup_{r \to 0} \frac{3}{4\pi r^3} \int_{\mathbb B(x,r)} \vert f \vert \, \mathrm d\lambda^3 \quad \lambda^3\text{-a.e.
}$$
Furthermore, we have $0 = \limsup_{r \to 0} \frac{4\pi}{3}r^2$, so we conclude 
$$ 0 = \left( \limsup_{r \to 0} \frac{4\pi}{3} r^2 \right)\left( \limsup_{r \to 0} \frac{3}{4\pi r^3} \int_{\mathbb B(x,r)} \vert f \vert \, \mathrm d\lambda^3 \right) \geq \limsup_{r \to 0} \int_{\mathbb B(x,r)} \vert f \vert \, \mathrm d\lambda^3 \quad \lambda^3\text{-a.e.} ,$$
so we conclude that 
$$\limsup_{r \to 0} \frac 1 r \int_{\mathbb B(x,r)} \vert f \vert \, \mathrm d\lambda^3 \; ,$$
which means, that $\lambda^3(A) = 0$. Is that correct?
At point (5) we set
$$\mathcal F := \left\{ \mathbb B(x,r) \, : \, \mathbb B(x,r) \subset U, \, x \in A_\sigma, \, 0 < r < \frac{\delta}{10}, \, \int_{\mathbb B(x,r)} \vert f \vert \, \mathrm d\lambda^3 > \sigma r \right\} \; .$$
Then we find by Vitali, a countable collection $\{ B_i \}_{i=1}^\infty$ of disjoint balls in $\mathcal F$ with $A_\sigma \subset \bigcup_{i=1}^\infty \hat{B_i}$, and we calculate
$$\mathcal H^1_\delta(A_\sigma) \leq \sum_{i=1}^\infty 5 r_i \leq \frac{5}{\sigma} \sum_{i=1}^\infty \int_{B_i} \vert f \vert \, \mathcal d\lambda^3 \leq \frac{5}{\sigma} \int_U \vert f \vert \, \mathcal d\lambda^3 \leq \frac{5\epsilon}{\sigma} \; .$$
The rest follows easely. Is that correct?
