Given a non-empty open set $\Omega \subset \mathbb R^n$, can one guarantee that $f = 0$ almost everywhere on $\Omega$ under a specific condition?

Context. Let $1 \leqslant p < \infty$ and $0 \leqslant \lambda \leqslant n$ be fixed elements. Moreover, suppose that $\Omega \subset \mathbb R^n$ is a non-empty open set. Throughout this post I will be dealing with the usual Lebesgue measure on $\mathbb R^n$ and the Lebesgue integral.

Question. Let $f \in L^p_{\operatorname{loc}}(\Omega)$. My goal is to prove the following implication.

$$ \sup_{x \in \Omega, \, r > 0} r^{-\lambda} \int_{B(x,r) \cap \Omega}|f(y)|^p \, dy = 0 \implies f = 0 \text{ almost everywhere on } \Omega. $$

My attempt. Assuming that the left-hand side above holds, I was able to prove that

$$ \int_{B(x,r) \cap \Omega} |f(y)|^p \, dy = 0, $$

for every $x \in \Omega$ and $r > 0$, with a simple reasoning by contradiction (I can provide details on this, if needed). This conclusion obviously implies that

$$ f= 0 \text{ almost everywhere on } B(x,r) \cap \Omega, $$

for every $x \in \Omega$ and $r > 0$. However, I don't think this is enough to conclude that $f = 0$ almost everywhere on $\Omega$ and at the same time I can't think about other conclusion that the left-hand side allows me to retrieve.

Thanks for any help in advance.