We know that the maximal operator is bounded on $L^{p}(\mathbb{R}^{n})$ where $n\geq 1$ and $1<p<\infty$ and the proof would be contained in many classical harmonic analysis books. Here I find a much more different method to verify this when $n=1$ and $p=2$ (This method would be found in the book "Some Topics in Dyadic Harmonic Analysis" by Michael T.Lacey, pp.13-14).

The general ideal is that since $L^{2}(\mathbb{R})$ is a Hilbert space, therefore, the bounded linear operator $T$ on $L^{2}(\mathbb{R})$ has the property $\left \| T \right \|^{2}\leq \left \| TT^{\star} \right \|$. Thus the essential thing is to construct a proper bounded linear operator $T$ on $L^{2}(\mathbb{R})$ such that $Mf(x)\leq CTf(x)$ where $C>0$ is a fixed constant and $f(x)$ is positive. This operator has been given by the form in Lacey's book: \begin{align*} Tf(x)=\sum_{I\in \mathfrak{D}}\frac{\chi_{E(I)}}{|I|}\int_{I}f(y)dy \end{align*} where $\mathfrak{D}$ denotes the set of dyadic intervals on $\mathbb{R}$ and for each $I\in \mathfrak{D}$ associate $E(I)\subseteq I$ so that the sets $\{E(I): I\in \mathfrak{D}\}$ are disjoint subsets. Lacey points out that $Mf(x)\leq 2Tf(x)$ in his book.(I could not verify this even if taking $C$ instead of $2$.)

To Bound the maximal operator, we calculate \begin{align*} TT^{\star}f(x)=\sum_{I\in \mathfrak{D}}\sum_{J\in \mathfrak{D}}\chi_{E(I)}\frac{<\chi_{I},\chi_{J}>}{|I|}\frac{<\chi_{E(J)},f>}{|J|}\leq Tf(x)+T^{\star}f(x). \end{align*} Therefore, from the inequalities \begin{align*} \left \| Tf \right \|^{2}_{2}\leq \left \| T \right \|^{2}\left \| f \right \|^{2}_{2}\leq \left \| TT^{\star} \right \|\left \| f \right \|^{2}_{2}\leq 2\left \| T \right \|\left \| f \right \|^{2}_{2} \end{align*} we have $\left \| T \right \|\leq 2$ and then $\left \| Mf \right \|_{2}\leq C\left \| Tf \right \|_{2}\leq 2C\left \| f \right \|_{2}$.

However, for the key step, the description of the choice of $E(I)$ in Lacey's book seems not too clear (maybe to me) . Hence, $\mathbf{my \ question}$ is:

How to choose a desirable set $\{E(I): I\in \mathfrak{D}\}$ such that the pointwise inequality $Mf(x)\leq CT(x)$ is valid？