Maximal functions where weak type inequality fails Let $\mathcal{R}$ denote the set of all open rectangles in $\mathbb{R}^2$ with sides parallel to the coordinate axis. Given a function on $\mathbb{R}^2$, consider the maximal function: 
$$f_{\mathcal{R}}^*(x)=\sup_{x\in R\in \mathcal{R}}\frac{1}{|R|}\int_R|f(y)|dy,\enspace x\in \mathbb{R}^2 ,$$
Problem the goal of this problem is to show that $f\mapsto f_{\mathcal{R}}^*$ does not satisfy the weak type inequality 
$$\left| \left\{x\in \mathbb{R}^2:\enspace  f_{\mathcal{R}}^*(x)>\alpha\right\} \right|\leq \frac{C}{\alpha}||f||_{L^1(\mathbb{R}^2)}, \forall \alpha>0, f\in L^1(\mathbb{R}^2)  $$
for some constant $C>0$ that is independent of $f$ and $\alpha$. 
This is done in 3 steps: 
Step1: Show that if $\{f_n\}_{n=1}^{\infty}$ is a sequence of non-negative measurable functions on $\mathbb{R}^d$, then for any $t>0$ we have: 
$$m\left(\left\{x\in \mathbb{R}^d:\enspace \liminf_{n\to\infty}f_n(x)>t\right\} \right)
\leq  \liminf_{n\to\infty}m\left(\left\{x\in \mathbb{R}^d:\enspace f_n(x)>t\right\} \right).$$
**Step 2: ** Let $B=B_1(0)$ denote the unit ball in $\mathbb{R}^2$ and let $\varphi(x)=\frac{1}{\pi}\chi_B(x)$. For $\delta>0$, set 
$$\varphi_{\delta}(x)=\delta^{-2}\varphi(x/\delta)=\frac{1}{|B_{\delta}(0)|}\chi_{B_{\delta}(0)}(x). $$
Prove for that $x=(x_1,x_2)\in \mathbb{R}^2$ such that $|x_1x_2|>0$,
$$\liminf_{\delta\to 0^+}(\varphi_\delta)_{\mathcal{R}}^*(x)\geq \frac{1}{16|x_1||x_2|}. $$
Step 3: Combine step 2 and 3 while assuming that the weak type inequality does hold to yield a contradiction.
Attempt
With some help here: Problem related to Fatou's Lemma (Measure theory) I was able to solve step 1, and I was able to get the estimate for step 2 also. Assuming the inequality and applying it to the map $\varphi_\delta$ for $\delta>0$, noting that $||\varphi_\delta||_1=1$, and then taking $\liminf$ as $\delta\to0^+$ and invoking the result from step 1 I was able to obtain 
$$\left|\left\{x: \enspace \liminf_{\delta\to0^+}(\varphi_{\delta})_{\mathcal{R}}^*(x)>\alpha \right\} \right| \leq \frac{C}{\alpha}$$
which in turn implies by step 2 that 
$$\left|\left\{(x_1,x_2)\in \mathbb{R}^2:\enspace |x_1x_2|>0 \enspace and \enspace \frac{1}{16|x_1||x_2|}>\alpha \right\} \right| \leq C/\alpha.$$
I believe at this point a contradiction can be obtained but I am not sure exactly how to show the measure of the set on the LHS can be "bigger". I have been told from this point that the LHS is of order $(\log \alpha)/ \alpha$ as $\alpha\to \infty$ which proves the claim, but I am not too sure what this means or how to prove it.
 A: Assume there exists such a $C>0$ (independent of $f$ and $\alpha$) s.t.
$$\left| \left\{x\in \mathbb{R}^2:\enspace  f_{\mathcal{R}}^*(x)>\alpha\right\} \right|\leq \frac{C}{\alpha}||f||_{L^1(\mathbb{R}^2)}, \forall \alpha>0, f\in L^1(\mathbb{R}^2)  .$$
Then as above, we can apply this with $f=\varphi_\delta$ ($\delta>0$), and using the inequalities from step 1 and step 2 we get the following inequality for 
$E:=\left\{(x_1,x_2)\in \mathbb{R}^2:\enspace |x_1x_2|>0 \enspace and \enspace \frac{1}{16|x_1||x_2|}>\alpha \right\}$:
$$|E|\leq C/\alpha\text{ for all }\alpha>0. $$
Note that $F:=\{(x_1,x_2)\in \mathbb{R}^2:0<x_1,x_2<\infty\text{ and }\frac{1}{16x_1x_2}>\alpha \}\subset E$ and so also then $|F|\leq C/\alpha\enspace (*)$. 
Then by Fubini's Theorem (or construction of product measure) we have:
$$|F|= \int_{(0,\infty)}|F^y|dy\enspace \text{ where }F^y=\{x\in \mathbb{R}:(x,y)\in F\}.$$
One checks directly that $F^y=\left\{x:0<x<\frac{1}{16y\alpha}\right\}\implies |F^y|=\frac{1}{16y\alpha}$. Thus we have 
$$|F|=\frac{1}{16\alpha}\int_{(0,\infty)}\frac{1}{y}dy\geq\frac{1}{16\alpha}\int_{(\alpha^{-1},1)}\frac{1}{y}dy $$
where here I have restricted $\alpha>1$. Evaluating this integral gives
$$|F|\geq \frac{1}{16\alpha}(\log 1-\log(\alpha^{-1}))=\frac{1}{16\alpha}(\log(\alpha)).$$
By $(*)$ we then have $\frac{1}{16\alpha}(\log(\alpha))\leq C/\alpha$ for all $\alpha>1$. Rearranging this forces $16C\geq \log(\alpha)$ for all $\alpha>1$
which of course is a contradiction since $C$ is a finite constant and $\log(\alpha)\to\infty$ as $\alpha\to \infty$.
Thus, there cannot exist a $C$ that allows for such a weak type inequality. 
