Prove or disprove a claim related to $L^p$ space The following question is just a toy model:
Let $f:[0,1] \rightarrow \mathbb{R}$ be Lebesgue integrable, and suppose that for any $0\le a<b \le1$, $$\int_a^b |f(x)|dx \le \sqrt{b-a}$$ then prove or disprove that $$ \sup \left\{\frac{\int_E |f|dx}{|E|^{1/2}}: E \subset [0,1]\right\}<+\infty$$

If the claim above is false, then is it possible to prove that for any fixed $0<t<1/2$,  $$ \sup \left\{\frac{\int_E |f|dx}{|E|^{t}}: E \subset [0,1]\right\}<+\infty$$

Motivation: This is a long story. Throughout the following, we assume that $f$ is a measurable function from a bounded regular open set $\Omega \subset \mathbb{R}^n$ to $\mathbb{R}$. 
We know that if $f$ is $L^p(\Omega) \space(p>1)$, then by Holder's inequality, $$ \sup \left\{\frac{\int_E |f|dx}{|E|^{1-1/p}}: E \subset \Omega\right\}<+\infty \quad \quad\quad\quad(*)$$Then naturally I wanted to ask the inverse question: 
$\space$ Does $(*)$ imply $f \in L^p(\Omega)$? 
One of my smart friends figured out that $(*)$ is equivalent to $f$ is weak $L^p$. See the answer here: a characterization of $L^p$ space. 
Then naturally I want to know in $(*)$, instead of taking supremum over all measurable sets, what would happen if taking supremum over all cubes or balls? This leads to the toy model I asked at the beginning: the toy model is for $p=2, n=1$, $f$ is integrable, and cube is thus just an interval.
Now let me formulate my question neatly as follows: Set
$M_p:=\left\{f:\sup \left\{\frac{\int_E |f|dx}{|E|^{1-1/p}}: E \subset \Omega\right\}<+\infty\right\}$ 
$\tilde{M_p}:=\left\{f:\sup \left\{\frac{\int_B |f|dx}{|B|^{1-1/p}}: \text{$B$ is a ball $\subset \Omega$}\right\}<+\infty\right\}$ 
$L_p^w$ := the weak-$L^p$ space.
$\tilde{L_p}:=\{f \in L^q(\Omega): \forall 1\le q<p\}$
Then by my friend's result and interpolation theorem, $$L_p^w=M_p \subset \tilde{L_p}$$Also trivially, $M_p \subset \tilde{M_p}$. So the ultimate goal is that I want to know the relationship between $M_p, \tilde{M_p}$, and $\tilde{L_p}$. In particular, the toy model I asked at the beginning focuses on whether $M_p = \tilde{M_p}$. 
An equivalent statement of whether $M_p=\tilde{M_p}$ is the following:
Let $0<s<1$. If $\mu$ is a finite measure on $\Omega$ and absolutely continuous with respect to Lebesgue measure in $\mathbb{R}^n$, and 
$\lim\sup _{r \rightarrow 0} \frac{\mu({B_r(x)})}{r^{ns}} \le 1, \forall x\in \Omega$ , is it true that $sup \{\frac{\mu(E)}{|E|^s}: E \subset \Omega\}<+\infty$ ?
I also want to understand the following question: 
$\space$ If $M_p = \tilde{M_p}$ and $f \in M_p$, is it true that $\sup \left\{\frac{\int_E |f|dx}{|E|^{1-1/p}}: E \subset \Omega\right\}=\sup \left\{\frac{\int_B |f|dx}{|B|^{1-1/p}}: \text{$B$ is a ball $\subset \Omega$}\right\}?$ Or what can we say about the ratio?
By the way, the definition $\tilde{M_p}$ here is the same as $M^p$ defined in Gilbarg and Trudinger on Page 164, which is the so called Morrey Space. I looked up some references but didn't find any claims whether or not $M^p \subset L^q \quad \forall 1 \le q < p$.
Maybe I'm thinking too much. I should focus on solving one problem and then go step by step.  

My effort: 
In terms of the possible approaches, I think the approximate continuity of any measurable function and a nice covering argument would be helpful. Also, if $f$ is integrable, then one can observe that if $$T_pf(x):=\lim\sup_{r \rightarrow 0} \frac{1}{|B_r(x)|^{1−1/p}}\int_{B_r(x)}|f(y)|dy$$ is bounded in $\Omega$, then $$\mathcal{M}_pf(x):=\sup_{r > 0} \frac{1}{|B_r(x)|^{1−1/p}}\int_{B_r(x)}|f(y)|dy$$ is also bounded, and vice versa. Also, $$T_pf(x)=0,\mathcal{H}^{s}-a.e, \forall s\ge 1-1/p$$ So the size of the blow-up points should be very small, and thus a nice covering argument may be applied, at least we don't need to worry about cover the singular sets by balls or other arbituary sets. Maybe at least $\tilde{M_p} \subset M_q, \forall 1\le q<p$ can be provable. I have a lot of other observations, but it is cumbersome to type them down. Overall, I think these problems should be related to geometric measure theory and are not trivial.
Also, my smart friend suggests me try to apply the Littlewood-Paley Theory. He thinks of them as standard problems in harmonic analysis.  

Any ideas, comments and partial result would be fully appreciated. I've no idea even about the toy model proposed.
 A: We want to mimic $f(x) = x^{-1/2}$ on $[0,1]$, but then cut $[0,1]$ into lots of intervals, and then on each interval remake the function so that its integral over that interval remains the same, but the support of the function is much smaller.
So let $x_n \to 0$ be a decreasing sequence with $x_0 = 1$.  Write $x_{n} = (1+\epsilon_n) x_{n+1}$, and suppose that $1 > \epsilon_n \to 0$ is a decreasing sequence.  Then define $f$ on $[x_{n+1},x_n)$ to be
$$ f(x) = \cases{\frac1{\epsilon_n} x_{n}^{-1/2} & if $x_{n+1} \le x \le (1 + \epsilon_n^2) x_{n+1}$ \cr 0 & if $ (1 + \epsilon_n^2) x_{n+1} \le x < (1 + \epsilon_n) x_{n+1}$\cr} $$
Check that $\int_a^b f(x) \, dx \le \sqrt{b-a}$.
Now set $E_N = \bigcup_{n=N}^\infty [x_{n+1},(1 + \epsilon_n^2) x_{n+1})$.  Then $\int_{E_N} f(x) \, dx = \int_0^{x_N} f(x) \, dx \approx \sqrt{x_N}$, whereas $|E_N| \le \epsilon_N x_N$.
Suppose $x_n = 1/n^\alpha$ for $0<\alpha \le 1$.  Then $\epsilon_n \approx \alpha/n$.  Hence
$$ \int_{E_N} f(x) \, dx \gtrsim (\alpha^{-1}|E_N|)^{\alpha/(2(\alpha+1))} .$$
Still lots of details to be checked.  But I think this will provide a counterexample.
A: Through out the following, I use $P(E)$ to denote the perimeter of the set $E$. (See the definition of "sets of finite perimeter" in Giusti, minimal surfaces and functions of bounded variation. One may think of it as surface area.)
The following example shows $h \in \tilde{M}_p$, but $h \notin M_q,\forall$ $\frac{1}{1-t}\le q<p$
Let $Q_k=[x_{k+1},x_k) \times [x_{k+1},x_k)$, and define $h$ on $Q_k$ to be $h_k=k^{1+\alpha}$, where $\alpha$ is to be specified later, and zero otherwise. Let $E_K=\bigcup_{k=K}^{\infty}Q_k$. 
Since $x_k=k^{-\alpha}$, $x_k-x_{k+1}\approx k^{-\alpha-1}$, then we have the following:
$|Q_k| \approx k^{-2\alpha-2}, P(Q_k) \approx k^{-1-\alpha},
\int_{Q_k} h\approx k^{-1-\alpha},
|E_K| \approx K^{-1-2\alpha}$ and$\int_{E_K}h \approx K^{-\alpha}$
For any $t \in (0,1/2)$, choose $0<\alpha < \frac{1}{1-2t}$, then
$$\frac{\int_{E_K}h}{|E_K|^t} \approx \frac{K^{-\alpha}}{(K^{-1-2\alpha})^t}=K^{1-\alpha(1-2t)}\rightarrow \infty$$
Therefore, since $M_p(\Omega) \subset M_q(\Omega), \forall 1 \le q \le p$ and let $p_0 = p_0(t)=\frac{1}{1-t}$, $h \notin M_p(\Omega)$ for any $p \ge p_0$. 
However, for any $E \subset \Omega$ and $E$ has finite perimeter, since $\mathcal{H}^1(Q_k \cap Q_{k-1})=\mathcal{H}^1(\{(x_k,y_k)\})=0$,we have $$P(E) \ge P(E \bigcap (\cup_{k=1}^{\infty} Q_k) = \Sigma_{k=1}^{\infty} P(E\cap Q_k)$$
Also, since $h$ is supported in $\cup_{k=1}^{\infty} Q_k$, we have $$\int_E h=\Sigma_{k=1}^{\infty}\int_{E\cup Q_k}h$$
Therefore, $$\frac{\int_E h}{P(E)} \le \frac{\int_E h}{\Sigma_{k=1}^{\infty} P(E\cap Q_k)} \le \sup\{\frac{\int_F h}{P(F)}: F \subset Q_k, k=1,2, \dots\}$$For any $F \subset Q_k$, we have by the isoperimetric inequality that $$\frac{\int_F h}{P(F)} = h_k |F|/P(F) \le C h_k P(F) \le C h_k P(Q_k) \approx k^{1+\alpha}k^{-1-\alpha}=1 \quad (1)$$where C in the estimate above is the isoperimetric constant in $\mathbb{R}^2$, thus $$\sup\{\frac{\int_F h}{P(F)}: F \subset Q_k, k=1,2, \dots\} < \infty \quad \quad $$ and thus  $$\sup\{\frac{\int_E h}{P(E)}: E \subset \Omega\} < \infty$$
By (1), we conclude that for any cube $Q \subset \Omega$,  $\frac{\int_Q h}{P(Q)} \le C $. Since $Q$ is a cube, $P(Q) \approx |Q|^{1/2}$, hence $\frac{\int_Q h}{|Q|^{1/2}} \le C $. Therefore, we showed that $$ \sup\{\frac{\int_Q h}{|Q|^{1/2}}: Q \subset \Omega\} \le C,$$ thus $h \in \tilde{M}_2$    

Remark: I did plan to summarize what I had done, and share some experiences, but I'm lack of time. So I just put another example here, whose proof is very sneaky. Maybe it's better to know this idea. Thanks to @Stephen Montgomery-Smith's inspiration, I found a whole bunch of examples by myself, and I only choose this one to post because it is quite interesting.
