Suppose that $f$ is a Lebesgue measurable function defined on an arbitrary open set $\Omega \subset \mathbb R^n$. Furthermore, assume that $f \in L^p_{\operatorname{loc}}(\Omega)$ and that
$$ \sup_{x \in \Omega, \, r > 0} r^{-\lambda} \int_{B(x,r) \cap \Omega} |f(y)|^p \, dy < \infty, $$
where $\lambda$ is an arbitrary real constant such that $\lambda > n$.
GOAL. My goal is to show that, under these conditions, it follows that $f = 0$ almost everywhere on $\Omega$.
Precedent work and why it fails. Severin Schraven has provived an answer to this problem that we thought to be good enough. After some time, while re-analysing this question, I figured out that there was a slight problem with his answer. The main result that he used was the Lebesgue Differentiation Theorem which is usually formulated as
Lebesgue Differentiation Theorem (LDT). Given a function $g \in L^1_{\operatorname{loc}}(\mathbb R^n),$ we have that
$$ \lim_{r \to 0} \frac{1}{|B(x,r)|} \int_{B(x,r)} g(y) \, dy = g(x) $$
for almost every $x \in \mathbb R^n$.
Schraven's answer was based on applying the LDT to $|f|^p \chi_\Omega$ but the problem is that with this version of the LDT we're unable to do this: note that $f \in L^p_{\operatorname{loc}}(\Omega)$ isn't enough to guarantee that $|f|^p \chi_\Omega \in L^1_{\operatorname{loc}}(\mathbb R^n)$. Therefore, I am looking for alternative aproaches to guarantee the desired result.
Thanks for any help in advance.
Best Answer
This really should be a comment, but it’s a little too long for it.
With your assumptions, let $M_{\lambda}$ denote that finite supremum. Now, fix any point $x\in\Omega$. Since $\Omega$ is open, there exists an $r_x>0$ such that $B(x,r_x)\subset\Omega$. Then, for any $0<r<r_x$, we have that $B(x,r)\cap \Omega= B(x,r)$. Letting $c_n$ denote the volume of the unit ball, we have \begin{align} \frac{1}{|B(x,r)|}\int_{B(x,r)}|f(y)|^p\,dy&=\frac{1}{c_nr^n}\int_{B(x,r)\cap \Omega}|f(y)|^p\,dy\leq\frac{1}{c_nr^n}\cdot M_{\lambda}r^{\lambda}=\frac{M_{\lambda}}{c_n}r^{\lambda-n}. \end{align} Since $\lambda>n$, we have that the LHS vanishes as $r\to 0^+$. Hence, by Lebesgue’s differentiation theorem, it follows that $f=0$ a.e on $\Omega$.
Notice I didn’t use the full strength of all your assumptions. In particular we can allow the $\lambda$ to depend on $x$, and we only need the $\limsup\limits_{r\to 0^+}$ to be finite for a.e $x\in \Omega$ (as opposed to the supremum over all $x$).
You’re right that if we’re in $L^1_{\text{loc}}(\Omega)$, it doesn’t necessarily imply we’re in $L^1_{\text{loc}}(\Bbb{R}^n)$, but it doesn’t matter because we’re considering integrals over smaller and smaller neighborhoods of a given point, so we can modify the function away from there.
Again, note that for all sufficiently small $r>0$, the sets $B(x,r)$ are contained in $\Omega$, so the integrals make sense. Also notice that I am referring to a specific function $f$ (not an equivalence class), because otherwise we cannot define the notion of Lebesgue point (since it uses the evaluation $f(x)$).
Let’s say you’ve already proved this theorem when $\Omega=\Bbb{R}^n$ and for $L^1(\Bbb{R}^n)$ functions. To deduce the theorem in the above case, we simply localize the argument: we can find a sequence $\{U_n\}_{n=1}^{\infty}$ of open sets such that each $\overline{U_n}$ is compact, and contained in $U_{n+1}$ and $\bigcup\limits_{n=1}^{\infty}U_n=\Omega$. For each $n$, let us define $f_n:\Bbb{R}^n\to\Bbb{C}$ to be $f$ on $U_n$ and $0$ elsewhere. Then, $f_n\in L^1(\Bbb{R}^n)$, so for a.e $\xi\in\Bbb{R}^n$, we have \begin{align} \lim_{r\to 0^+}\frac{1}{|B(\xi,r)|}\int_{B(\xi,r)}|f_n(y)-f_n(\xi)|\,dy&=0. \end{align} In particular this equality holds for a.e $x\in U_n$, i.e there is a null set $Z_n\subset U_n$ such that for all $x\in U_n\setminus Z_n$, the above equality holds. But notice that since $x\in U_n$ and $U_n$ is open, for small enough $r>0$, we’ll have $B(x,r)\subset U_n$ and so the integrand above will become $f$ instead of $f_n$. In other words \begin{align} \lim\limits_{r\to 0^+}\frac{1}{|B(x,r)|}\int_{B(x,r)}|f(y)-f(x)|\,dy&=0.\qquad\text{($x\in U_n\setminus Z_n$)} \end{align} Since this is true for all $n$, we can upgrade the the above equality and deduce it holds for all $x\in \bigcup\limits_{n=1}^{\infty}(U_n\setminus Z_n)=\Omega\setminus\left(\bigcup_{n=1}^{\infty}Z_n\right)$, which is $\Omega$ minus a countable union of measure-zero sets (which is still measure zero). Thus, we have shown that a.e point of $\Omega$ is a Lebesgue-point of $f$.