8
$\begingroup$

Let $f(t)$ be a measurable and almost everywhere finite function, defined on the closed interval $E = [a, b]$. Prove the existence of a decreasing function $g (t)$, defined on [a, b], which satisfies the relation $m(E \cap \left \{ x: {g > x} \right \}) = m(E \cap \left \{ x: {f > x} \right \}) $ for all real x.

I'm not sure I can use the theorem:

Let $f(x)$ be defined on the entire real line $\mathbb{R}$ and be measurable and finite almost everywhere. Then, for every $\varepsilon > 0$ and $\alpha> 0$, there exists a continuous function $g (x)$ defined on the entire line such that $m(\mathbb{R}\cap x:|f-g| \geq \alpha)< \varepsilon$. Thanks.

$\endgroup$
  • 2
    $\begingroup$ The theorem tells that you can approximate $f$ by continuous functions uniformly except at a set of small measure. On the other hand, the function $g$ you are asked to construct in the question is by no means an approximation of $f$. It is more like 'slicing the area under the graph horizontally and pushing the slices to the left'. This would be much easier to imagine if you know Cavalieri's principle. $\endgroup$ – Sangchul Lee Sep 22 '17 at 21:02
  • 2
    $\begingroup$ I guess the equality between measures should be $m(E \cap \left \{ t: {g(t) > x} \right \}) = m(E \cap \left \{ t: {f(t) > x} \right \})$ instead. $\endgroup$ – Alex Ravsky Sep 23 '17 at 7:11
  • $\begingroup$ If we don’t allow infinite values of the function $g$ then the claim is false. Indeed, let $[a,b]=1$ and $f(x)=0$, if $x=0$ and $f(x)=1/x$, otherwise. Then $m(\left \{ t: {g(t) > g(0)} \right \})=0\ne m(\left \{ t: {f(t) > g(0)} \right \}),$ a contradiction. $\endgroup$ – Alex Ravsky Sep 24 '17 at 15:04
  • 1
    $\begingroup$ I guess there can be something like $g(y)=\sup\{x: m(\{ t\in E: {f(t) > x}\})\ge y\}$ for each $y\in E$. $\endgroup$ – Alex Ravsky Sep 24 '17 at 15:18
  • 2
    $\begingroup$ Possibly related: en.wikipedia.org/wiki/Symmetric_decreasing_rearrangement $\endgroup$ – Bungo Sep 25 '17 at 5:19
3
+25
$\begingroup$

Let's begin by assuming $f$ is simple, i.e. of the form $\sum\limits_{n=1}^N a_nX_{A_n}$ with $\{a_n\}_{n=1}^N\subset \mathbb{R}$ and $\{A_n\}_{n=1}^N\subset \mathcal{B}_{[a,b]}$ with $A_i\cap A_j=\emptyset$ for $i\neq j$. Then we can WLTOG assume that $\{a_n\}_{n=1}^N$ is a decreasing sequence. We see that $f: [a,b] \rightarrow \mathbb{R}$ defined by $f=a_1X_{[a,\text{ }a_2+\mu(A_1)]}+a_2X_{[a+\mu(A_1),\text{ }a_2+\mu(A_1)+\mu(A_2)]}+ \cdots a_nX_{[a+\sum_{n=1}^{N-1}\mu(A_n),\text{ }a+\sum_{n=1}^N\mu(A_n)]}$ is non-increasing and that $\mu(f^{-1}((x,\infty)))=\mu(g^{-1}((x,\infty)))$ for any $x\in \mathbb{R}$.

We now proceed by bootstrapping. If $f$ is non-negative we know there is a non- decreasing sequence of simple functions, $\{s_n\}_{n=1}^{\infty}$, defined on $[a,b]$ so that $\{s_n\}_{n=1}^{\infty}$ converges to $f$ uniformly. It is then quite clear that $\mu(f^{-1}((x,\infty)))=\lim\limits_{n \rightarrow \infty} \mu({s_n}^{-1}((x,\infty)))$ for any $x \in \mathbb{R}$ (1). We know from the above paragraph that for any $n\in\mathbb{N}$ there exists non-increasing $g_n$ so that $\mu({s_n}^{-1}((x,\infty)))=\mu({g_n}^{-1}((x,\infty)))$ for all $x\in \mathbb{R}$ (2). $\{g_n\}_{n=1}^{\infty}$ is a non- decreasing sequence and so converges pointwise in $\mathbb{R}^*$. Thus we can define $g=\lim\limits_{n \rightarrow \infty} g_n$ which must also be non-increasing.

As $g$ dominates $\{g_n\}_{n=1}^{\infty}$ it is clear that given $x\in \mathbb{R}$, $\mu(g^{-1}((x,\infty)))\geq \mu({g_n}^{-1}((x,\infty)))$ for all $n$ so $\mu(g^{-1}((x,\infty)))\geq \lim\limits_{n \rightarrow \infty} \mu({g_n}^{-1}((x,\infty)))$. It remains to show the reverse inequality so let $\epsilon>0$ be given. If $g^{-1}((x, \infty))=\emptyset$ there is nothing to check so assume it isn't. Set $c=\sup(g^{-1}((x, \infty))-\epsilon$, then $g(c)>x$ and as $g_n\uparrow g$ pointwise there is a $N$ such that $g_n(c)>x$ for $n\geq N$. It follows that $[a,c]\subset {g_n}^{-1}((x,\infty))$ and then by the construction of $c$ $\mu(g^{-1}((x,\infty)))-\epsilon=\mu([a,c])\leq\mu({g_n}^{-1}((x,\infty)))$. It follows that $\mu(g^{-1}((x,\infty)))-\epsilon\leq \lim\limits_{n \rightarrow \infty} \mu({g_n}^{-1}((x,\infty)))$. As $\epsilon>0$ was arbitrary we have shown that $\mu(g^{-1}((x,\infty)))=\lim\limits_{n \rightarrow \infty} \mu({g_n}^{-1}((x,\infty)))$. Combine this with 1 and 2 and we have $\mu(f^{-1}((x,\infty)))=\mu(g^{-1}((x,\infty)))$ for all $x\in \mathbb{R}$.

We can now attempt one final push to extend the result to a measurable $f$ that is finite a.e. For such $f$ define the non- increasing sequence $\{f_n\}_{n=1}^{\infty}$, by $f_n=fX_{f^{-1}([-n,\infty))}$. These functions aren't necessarily non-negative, but adding $n$ to them makes them non-negative so by the previous paragraph it should be clear that for all $n$ there exists a non-increasing function $g_n$ such that $\mu({f_n}^{-1}((x,\infty)))=\mu({g_n}^{-1}((x,\infty)))$. $\{g_n\}_{n=1}^{\infty}$ is of course a non-increasing sequence itself so we can define $g=\lim\limits_{n \rightarrow \infty} g_n$. This function must be non-increasing. By pretty much the same argument as in the previous paragraph, $\mu(g^{-1}((x,\infty)))=\lim\limits_{n \rightarrow \infty} \mu({g_n}^{-1}((x,\infty)))$. Because $f$ and $f_n$ agree except on sets of measure shrinking to $0$ (here we are using that the set on which $f$ takes on the value $-\infty$ has measure $0$) we also have $\mu(f^{-1}((x,\infty)))=\lim\limits_{n \rightarrow \infty} \mu({f_n}^{-1}((x,\infty)))$ for all $x\in \mathbb{R}$. So $\mu(f^{-1}((x,\infty)))=\mu(g^{-1}((x,\infty)))$ for all $x\in \mathbb{R}$.

Remark: in the above construction of $g$ we might get a function that attains the value $\infty$ at $a$ and/or $-\infty$ at $b$. We also see that since $f$ is finite a.e. $g$ cannot equal either of these values on $(a,b)$.

$\endgroup$
1
$\begingroup$

The comments and the cited wiki page on decreasing rearrangement give an expression for $g$ but little hints as to why it works. In the following I will try to provide an explanation.

Assuming $f:[a,b]\rightarrow [c,d]$, we define first the repartition function $F: [c,d] \rightarrow [a,b]$: $$ F(x) = a + m \{t\in [a,b] : f(t) >x \}, \; \; x\in [c,d].$$ (I have added $a$ to the result to map into the interval $[a.b]$). $F$ is monotone decreasing and, what is crucial here, right-continuous, since by $\sigma$-additivity of the measure: $$ F(x^+)-F(x)=\lim_{n\rightarrow \infty} F(x+\frac{1}{n}) -F(x) = \lim_{n\rightarrow \infty} m \{t\in [a,b] : x+\frac{1}{n}\geq f(t) >x \} = 0 .$$

The goal is to construct a monotone decreasing function, $g: [a,b]\rightarrow [c,d]$, having the same partition function as $f$: $$F(x) = a+ m\{ t\in [a,b]: g(t)>x\}, \; \; x\in [c,d].$$ Note first that when $g$ is decreasing, $g(t)>x$ implies $g(s)>x$ for every $s\in [a,t]$ so the above condition is equivalent to: $$ F(x) = \sup \{ t\in [a,b]: g(t)>x\}.$$ In the case that $F$ is a bijection from $[c,d]$ onto $[a,b]$, we may as $g$ in the above just take the inverse of $F$. Problems arise, however, when $F$ is not continuous or not injective. We claim that $g$ in any case may be constructed in the completely symmetric fashion: $$g(t) := \sup \{ x\in [a,b]: F(x)>t\}, \; \; t\in [a,b].$$ One may verify by $\epsilon$-$\delta$ argument that it satifies the wanted relation but the logic may (in my opinon) be difficult to digest for human beings.

So instead let me describe a different conceptual approach, in which the use of right-continuity will appear naturally: Write $R=[a,b]\times [c,d]$ and define $$ \Omega = \{ (t,x) \in R : F(x) \leq t \}$$ It is a subset of the rectangle having the properties:

(1) It is directed, i.e.: $ (t,x)\in \Omega \Rightarrow [t,b]\times [x,d]\subset \Omega$. (Make a drawing!)

(2) It is a closed subset of $R$. (This follows from the right-continuity of $F$).

Conversely, given a directed closed subset $\Omega$ of $R$ (i.e. having the above two properties) we may reconstruct a corresponding $F$ (which becomes right-continuous because $\Omega$ is closed) by setting: $$ F(x) = \sup\{t\in [a,b] : (t,x)\notin \Omega\}.$$ There is thus a bijection between a closed directed set $\Omega$ and a corresponding decreasing right-continuous map $F: [c,d]\rightarrow [a,b]$.

The properties of the set $\Omega$ is, however, completely symmetric in the way the two coordinates are treated. So it is equally well in unique correspondance with a function $g: [a,b]\rightarrow [c,d]$ where coordinates have been exchanged and which therefore may be constructed through: $$g(t) = \sup\{ x\in [c,d] : (t,x) \notin \Omega \} = \sup\{ x\in [c,d] : F(x)>t \} $$ Because of the symmetry, the procedure for going from $g$ to $F$ is the very same with the variables exchanged, i.e.: $$F(x) = \sup \{ t\in [a,b] : g(t)>x\} $$ But this was precisely the wanted property of the partition function. Magic?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.