[proofplan] The hypothesis $f \in L^1(\mathbb{R}^n)$ controls the total mass of $|f|$, while the volume $|Q_k| = 2^{-kn}$ blows up as $k \to -\infty$. The average is bounded above by $\|f\|_{L^1}/|Q_k|$, which vanishes purely from the volume growth. The proof formalises this single estimate, taking care that the integral domain enlargement and the measure normalisation are explicit. [/proofplan] [step:Bound the average by the global $L^1$ mass divided by the cube volume] Fix $f \in L^1(\mathbb{R}^n)$, a point $x \in \mathbb{R}^n$, and a sequence of dyadic cubes $(Q_k)_{k \in \mathbb{Z}}$ with $Q_k \in \mathcal{D}$ of generation $k$ and $x \in Q_k$. Recall that the generation-$k$ dyadic cube has side length $2^{-k}$, so \begin{align*} |Q_k| = \mathcal{L}^n(Q_k) = 2^{-kn}. \end{align*} Since $|f|: \mathbb{R}^n \to [0, \infty)$ is non-negative and $Q_k \subseteq \mathbb{R}^n$, monotonicity of the integral with respect to the integration domain (we enlarge $Q_k$ to $\mathbb{R}^n$) yields \begin{align*} \int_{Q_k} |f|\, d\mathcal{L}^n(y) \le \int_{\mathbb{R}^n} |f|\, d\mathcal{L}^n(y) = \|f\|_{L^1(\mathbb{R}^n)}. \end{align*} Dividing by $|Q_k| > 0$, \begin{align*} \frac{1}{|Q_k|}\int_{Q_k} |f|\, d\mathcal{L}^n(y) \le \frac{\|f\|_{L^1(\mathbb{R}^n)}}{|Q_k|} = \frac{\|f\|_{L^1(\mathbb{R}^n)}}{2^{-kn}} = 2^{kn}\,\|f\|_{L^1(\mathbb{R}^n)}. \end{align*} [/step] [step:Send the side length to infinity to obtain the limit] As $k \to -\infty$, $|Q_k| = 2^{-kn} \to \infty$. Since $\|f\|_{L^1(\mathbb{R}^n)} < \infty$ is a fixed finite constant by hypothesis, the upper bound from the previous step satisfies \begin{align*} 0 \le \frac{1}{|Q_k|}\int_{Q_k} |f|\, d\mathcal{L}^n(y) \le \frac{\|f\|_{L^1(\mathbb{R}^n)}}{|Q_k|} \xrightarrow[k \to -\infty]{} 0. \end{align*} The squeeze theorem applied to the lower bound $0$ and the upper bound $\|f\|_{L^1}/|Q_k|$ gives \begin{align*} \lim_{k \to -\infty} \frac{1}{|Q_k|}\int_{Q_k} |f|\, d\mathcal{L}^n(y) = 0, \end{align*} which is the claim. [/step]