[proofplan] The construction is a stopping-time argument on the dyadic grid. Call a dyadic cube **selected** if its average of $|f|$ exceeds $\lambda$ but no strict dyadic ancestor has this property — these are maximal cubes of average excess. The [Vanishing Average at Large Scales](/theorems/3153) result rules out selecting arbitrarily large ancestors, so maximality is well-defined; the [Lebesgue Differentiation Theorem](/theorems/???) ensures that points with $|f(x)| > \lambda$ are eventually inside a selected cube. Disjointness comes from the [Nesting Property of Dyadic Cubes](/theorems/???) combined with maximality. The average upper bound $2^n\lambda$ comes from comparing each selected cube to its dyadic parent (which was not selected). The measure bound is a direct consequence of the lower average bound $\lambda < \frac{1}{|Q_j|}\int_{Q_j}|f|\, d\mathcal{L}^n$ summed over the disjoint cubes. [/proofplan] [step:Define the selected dyadic cubes via a stopping-time on the average of $|f|$] For a dyadic cube $Q \in \mathcal{D}$, write \begin{align*} A(Q) := \frac{1}{|Q|}\int_Q |f|\, d\mathcal{L}^n(y) \end{align*} for the average of $|f|$ over $Q$. Call $Q \in \mathcal{D}$ **selected** if $A(Q) > \lambda$ and $A(Q') \le \lambda$ for every strict dyadic ancestor $Q' \in \mathcal{D}$ with $Q \subsetneq Q'$. Let \begin{align*} \mathcal{S} := \{Q \in \mathcal{D} : Q \text{ is selected}\} \end{align*} and enumerate $\mathcal{S} = \{Q_j\}_{j \in J}$ for some countable index set $J$ (the dyadic grid is countable, so $\mathcal{S}$ is countable). Set $\Omega := \bigcup_{j \in J} Q_j$. [/step] [step:Verify the selected cubes are pairwise disjoint via the dyadic nesting property] We claim the selected cubes $\{Q_j\}_{j \in J}$ are pairwise disjoint. Suppose $Q_i, Q_j \in \mathcal{S}$ with $Q_i \cap Q_j \neq \varnothing$. By the [Nesting Property of Dyadic Cubes](/theorems/???), either $Q_i \subseteq Q_j$ or $Q_j \subseteq Q_i$. Without loss of generality $Q_i \subseteq Q_j$. If $Q_i \subsetneq Q_j$, then $Q_j$ is a strict dyadic ancestor of $Q_i$ with $A(Q_j) > \lambda$ (since $Q_j$ is selected), violating the maximality clause for $Q_i$. Hence $Q_i = Q_j$. [/step] [step:Show $\Omega$ is open and $|f| \le \lambda$ a.e. off $\Omega$] The set $\Omega = \bigcup_{j \in J} Q_j$ is a union of half-open dyadic cubes, hence Borel measurable. To see it is open, note that a half-open cube $Q_{k,m} = \prod_{i=1}^n [2^{-k}m_i, 2^{-k}(m_i+1))$ is not itself open in the Euclidean topology, but $\Omega$ as a union of dyadic cubes from possibly different generations may be. We argue differently. Define \begin{align*} \Omega' := \{x \in \mathbb{R}^n : \exists\, Q \in \mathcal{D} \text{ with } x \in Q,\ A(Q) > \lambda\}. \end{align*} We show $\Omega = \Omega'$ up to a $\mathcal{L}^n$-null set, and that $\Omega'$ is open. For $x \in \Omega'$, pick $Q \in \mathcal{D}$ with $x \in Q$ and $A(Q) > \lambda$; among all dyadic ancestors of $Q$ — including $Q$ itself — by [Vanishing Average at Large Scales](/theorems/3153), only finitely many have average $> \lambda$ (since the average tends to $0$). Take the largest such ancestor $\tilde Q$: then $\tilde Q$ is selected, and $x \in \tilde Q \subseteq \Omega$. Hence $\Omega' \subseteq \Omega$. The reverse inclusion is immediate from the definitions. For openness: replace each half-open dyadic cube by its interior $\operatorname{int}(Q_j)$. The boundary $\partial Q_j$ consists of axis-parallel hyperplanes, which form a $\mathcal{L}^n$-null set. The set $\Omega^\circ := \bigcup_j \operatorname{int}(Q_j)$ is a union of open cubes, hence open, and $\Omega \setminus \Omega^\circ \subseteq \bigcup_j \partial Q_j$ has measure zero. We henceforth use $\Omega^\circ$ in place of $\Omega$ when openness is needed; the measure-theoretic statements are unaffected. Now we prove the off-set bound. By the [Lebesgue Differentiation Theorem](/theorems/???), for $\mathcal{L}^n$-a.e. $x \in \mathbb{R}^n$, \begin{align*} \lim_{k \to \infty} \frac{1}{|Q_{k,m_k(x)}|}\int_{Q_{k,m_k(x)}} |f|\, d\mathcal{L}^n(y) = |f(x)|, \end{align*} where $Q_{k, m_k(x)}$ is the unique generation-$k$ dyadic cube containing $x$ (the dyadic version of the differentiation theorem applies since dyadic cubes containing $x$ form a regular shrinking family). Take any $x \notin \Omega$ where this limit holds. By definition of $\Omega$, no dyadic cube containing $x$ is selected, so for every $k \in \mathbb{Z}$, the cube $Q_{k,m_k(x)}$ is not selected. We claim $A(Q_{k,m_k(x)}) \le \lambda$ for every $k$. Indeed, if $A(Q_{k,m_k(x)}) > \lambda$, then taking the maximal ancestor $\tilde Q$ of $Q_{k,m_k(x)}$ with $A(\tilde Q) > \lambda$ (which exists and is finite by [Vanishing Average at Large Scales](/theorems/3153)) would put $x \in \tilde Q \in \mathcal{S}$, contradicting $x \notin \Omega$. Letting $k \to \infty$, \begin{align*} |f(x)| = \lim_{k \to \infty} A(Q_{k,m_k(x)}) \le \lambda. \end{align*} This holds for $\mathcal{L}^n$-a.e. $x \notin \Omega$, proving (i). [/step] [step:Verify the lower average bound $A(Q_j) > \lambda$ from selection] By definition of "selected", every $Q_j \in \mathcal{S}$ satisfies $A(Q_j) > \lambda$, i.e., \begin{align*} \lambda < \frac{1}{|Q_j|}\int_{Q_j} |f|\, d\mathcal{L}^n(y). \end{align*} This is the lower bound in (ii). [/step] [step:Compare each selected cube to its dyadic parent to obtain $A(Q_j) \le 2^n \lambda$] Let $Q_j \in \mathcal{S}$ and let $\tilde Q_j \in \mathcal{D}$ be its unique dyadic parent: $\tilde Q_j \supsetneq Q_j$ with $|\tilde Q_j| = 2^n |Q_j|$. By maximality of $Q_j$, the parent $\tilde Q_j$ is not selected, and since $\tilde Q_j$ is not contained in any other selected cube (it would have to equal that cube by nesting and maximality, which is impossible since $\tilde Q_j \supsetneq Q_j$), the parent fails the average condition: $A(\tilde Q_j) \le \lambda$. Hence \begin{align*} \int_{Q_j} |f|\, d\mathcal{L}^n(y) \le \int_{\tilde Q_j}|f|\, d\mathcal{L}^n(y) = |\tilde Q_j|\, A(\tilde Q_j) \le 2^n |Q_j|\, \lambda, \end{align*} where we used $Q_j \subseteq \tilde Q_j$ and $|f| \ge 0$ to enlarge the integration domain. Dividing by $|Q_j| > 0$, \begin{align*} A(Q_j) = \frac{1}{|Q_j|}\int_{Q_j}|f|\, d\mathcal{L}^n(y) \le 2^n \lambda. \end{align*} [/step] [step:Sum the lower bound over the disjoint cubes to derive the measure estimate] For each $j$, the lower bound from the previous-but-one step gives $\lambda |Q_j| < \int_{Q_j}|f|\, d\mathcal{L}^n(y)$. Since the cubes $\{Q_j\}_{j \in J}$ are pairwise disjoint and $|f| \ge 0$, \begin{align*} \lambda \sum_{j \in J} |Q_j| < \sum_{j \in J} \int_{Q_j}|f|\, d\mathcal{L}^n(y) = \int_{\bigcup_j Q_j} |f|\, d\mathcal{L}^n(y) \le \int_{\mathbb{R}^n}|f|\, d\mathcal{L}^n(y) = \|f\|_{L^1}. \end{align*} The first equality uses countable additivity of the integral on disjoint measurable sets (a special case of the [Monotone Convergence Theorem](/theorems/???) applied to the partial sums $\sum_{j \le N} |f|\,\mathbb{1}_{Q_j}$, which increase pointwise to $|f|\,\mathbb{1}_\Omega$). The last inequality enlarges the integration domain from $\Omega$ to $\mathbb{R}^n$, valid because $|f| \ge 0$. Dividing by $\lambda > 0$, \begin{align*} \sum_{j \in J} |Q_j| \le \frac{\|f\|_{L^1}}{\lambda}, \end{align*} which is (iii). Together with the disjointness, the openness (modulo a null set), the off-set bound, and the two-sided average bound, this establishes (i)-(iii). [/step]