Measure Theory/Markov and Hardy-Littlewood

Markov's Inequality

We've already had enough preamble for this theorem. So let's just jump in and state, then prove, the following theorem.

Theorem: Let $f:ℝ\to ℝ$ be an integrable function and $c\in {ℝ}^{+}$.

Define $E_c=\{x\in ℝ:c\le |f(x)|\}$. (Think of this as the set of points at which f is large.)

Then

$\lambda (E_c)\le \frac{1}{c}\int f$

Prove Markov's inequality by construing $\lambda (E_c)$ as the integral of ${\mathrm{𝟏}}_{E_c}$, and then apply the ML bound.

(Would that it were so simple to prove the Hardy-Littlewood inequality.)

Hardy-Littlewood

Let $E_c=\{x\in ℝ:c<f^{*}(x)\}$ and we want to show that

$\lambda (E_c)<\frac{3}{c}\int f$

To start on the proof, we first observe that if $x\in E_c$ is arbitrary then there is some $t\in {ℝ}^{+}$ such that $\frac{1}{2t}\underset{(x-t,x+t)}{\int }|f|>c$.

This allows us to define, for each x, the corresponding open interval $(x-t,x+t)$ with t as above. This looks like, perhaps, a cover of $E_c$ by open intervals! That sounds provocative and familiar.

However, it would be senseless to cover for all of $E_c$, since there is no guarantee that $E_c$ is compact. Compactness is precisely what would make an open interval cover useful.

Whence we let $F\subseteq E_c$ be any compact subset. The idea behind what we will do for the remainder of this proof, is to show that $\lambda (F)\le \frac{3}{c}\int |f|$. It will then follow that $\lambda (E_c)\le \frac{3}{c}\int |f|$, which you will prove in an exercise below.

Now for each $x\in F$ we define the corresponding open interval $I_x=(x-t,x+t)$ such that $\frac{1}{2t}\underset{I_x}{\int }{f}^{*}>c$. By the compactness of F, there is a finite subcover, which we will choose to call $J_1,\dots ,J_n$.

We would like to reason as follows (although, of course, I would only phrase it this way if there is an obstacle coming):

$\begin{array}{cc}\lambda (F)& \le \sum _{i=1}^n\lambda (J_i)\\ & =\sum _{i=1}^{n}2t_i\text{ where }{t}_i\text{ is the width of interval }{J}_i\\ & <\sum _{i=1}^n\frac{1}{c}\underset{J_i}{\int }|f|\\ & \le \frac{1}{c}\int |f|\end{array}$

If this argument were correct then we could get a smaller bound on $\lambda (F)$, using $\frac{1}{c}\int |f|$ rather than $\frac{3}{c}\int |f|$.

However, the last inequality is not justified because the intervals $J_1,\dots ,J_n$ may overlap.

In this exercise you will justify each of the steps in

$\begin{array}{cc}\lambda (F)& \le \sum _{i=1}^n\lambda (J_i)\\ & =\sum _{i=1}^{n}2t_i\text{ where }{t}_i\text{ is the width of interval }{J}_i\\ & <\sum _{i=1}^n\frac{1}{c}\underset{J_i}{\int }|f|\\ & \le \frac{1}{c}\int |f|\end{array}$

except for the last one, which we have observed is actually invalid.

1. Explain why $\lambda (F)\le \sum _{i=1}^n\lambda (J_i)$.

2. Explain why $\lambda (J_i)=2t_i$.

3. Explain why $2t_i<\frac{1}{c}\underset{J_i}{\int }|f|$. Hint: Recall the defining property of $J_i$.

The Vitali Covering Lemma

In order to overcome the obstacle above, that $J_1,\dots ,J_n$ may fail to be disjoint, we will try to relate this collection to some other collection which is disjoint.

One strategy would be to simply merge overlapping intervals. For instance, if $J_1,J_2$ overlap each other, we could replace the pair with $J^{\prime }=J_1\cup J_2$. Repeating the procedure finitely many times would produce a new family which is now composed of disjoint intervals.

However, notice that if we did so, the we would no longer be able to say $\lambda (J_i)=2t_i$.

So we have two competing needs: The need for the intervals to be disjoint but also the need for the intervals to maintain their size.

The easiest resolution is to take the interval in $J_1,\dots ,J_n$ with the greatest length (ties may be broken arbitrarily), assume without loss of generality that this is $J_1$. If this intersects any other interval then we simply remove those intersecting intervals.

Now define $3*J_1$ to be the interval with the same center as $J_1$, but with three times its length.

Let $I=(a,b),J=(c,d)$ be two open bounded intervals. Assume that $\ell (I)<\ell (J)$ and that they intersect, $I\cap J\ne \mathrm{\varnothing }$.

Prove that $I\subseteq 3*J$.

Let $I_1,\dots ,I_m$ be any family of open bounded intervals. Show that there exists a sub-family $I_{k_1},\dots ,I_{k_n}$ with the properties

• the family is pairwise disjoint, and
• $\bigcup _{i=1}^m{I}_i\subseteq \bigcup _{i=1}^{n}3*I_{k_i}$

This sub-family is called the Vitali-covering for the family $I_1,\dots ,I_m$

The End

Using the Vitali-covering for the family $J_1,\dots ,J_n$ as in the initial "false proof", correct the false proof to obtain a correct proof that

$\lambda (F)\le \frac{3}{c}\int |f|$

and then conclude the theorem.

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