Before proceeding, I should introduce some context: during the post, I will denote the Lebesgue measure of a set $A$ by $|A|$. Furthermore, I define the set of dyadic cubes $\Delta$ by
$$ \Delta = \bigcup_{k \in \Bbb Z} \Delta_k, $$
where $\Delta_k$ is the set of all translates of the cube $Q_k = [0,2^{-k})^n$ whose vertices are on the lattice $2^{-k}\Bbb Z^n$, for each $k \in \Bbb Z$. Now, I introduce the definition of maximal dyadic operator I am working with:
Definition. Given a function $f \in L^1_{\text{loc}}(\Bbb R^n),$ define the dyadic maximal operator $M^d$ by
$$ M^d f(x) = \sup_{\substack{Q \ni x \\[.1cm] Q \in \Delta}} \frac{1}{|Q|}\int_Q |f(y)| \, dy. $$
The question itself. Now, consider the following set:
$$ E_t^d = \{x \in \Bbb R^n \, \colon \, M^df(x) > t\}, \quad \text{ for each } t > 0.$$
The book I am reading states the following (one should suppose that $E_t^d$ is not empty in the statement below):
Take $x \in E_t^d$. By the definition of the dyadic maximal operator, there exists $Q \in \Delta$ such that $x \in Q$ and $$ \frac{1}{|Q|}\int_Q |f(y)| \, dy > t. $$
And I am having quite a hard time seeing why this holds, altought I think this must be a trivial result.
Thanks for any help in advance.
This is from the definition of supremum. The maximal function is the supremum of all averages $\frac1{|Q|}\int_Q |f(y)|dy$, so if the supremum is greater than $t$ there must exist a cube whose average is greater than $t$. Indeed, if all cubes containing $x$ had corresponding average $\leq t$, then the supremum must be $\leq t$ since it is the least upper bound, a contradiction.