On continuity of distribution function and decreasing rearrangement

Abstract

Given a measure space $(\Omega,\Sigma,\mu)$ , the distribution function $\mu_{f}(\nu)=\mu(\{t\in\Omega:|f(t)|>\nu\})$ where $\nu\geqslant 0$ and the decreasing rearrangement $f^{*}(z)=\inf\{\nu\geqslant 0:\mu_{f}(\nu)\leqslant z\}$ , where $z\geqslant 0$ and by convention $inf\{\emptyset\}=\infty$ , of a measurable function $f$ are known to be right continuous functions. However, these functions need not be left continuous. The purpose of this paper is to investigate the conditions under which these functions are continuous. Under the assumption that $\mu(\{t\in\Omega:|f(t)|>0\})<\infty$ , we provide a necessary and sufficient condition for the function $\mu_{f}$ to be continuous at $\nu>0$ . Using the same we provide a similar result for the continuity of decreasing rearrangement $f^{*}$ of the function $f$ .

Keywords

Continuity distribution function decreasing rearrangement

1. Introduction

For any given measurable function $f$ on a measure space, the distribution function of $f$ describes certain properties of $f$ as well as the measure on the space. The decreasing rearrangement of $f$ is a non-negative function on $[0,\infty)$ , the measure of whose strict superlevel sets coincides with the measure of strict superlevel sets of $f$ . It is well known that both the distribution function and the decreasing rearrangement of $f$ are non-increasing and right continuous functions.

In this paper we study the continuity of these two functions namely the distribution function and the decreasing rearrangement of a given function. Our aim is to develop a method to find the points of continuity (or discontinuity) and provide necessary and sufficient conditions for continuity of these functions. We first quote few notations and definitions that are used in the sequel.

Let $(\Omega,\Sigma,\mu)$ be a measure space and $f:\Omega\to\mathbb{R}$ be a measurable function. The strict superlevel set of $f$ at point $\nu\in\mathbb{R}$ is denoted by $L(f,\nu)$ and defined by $\{t\in\Omega:f(t)>\nu\}$ . The distribution function of $f$ , denoted by $\mu_{f}$ , is a map $\mu_{f}:[0,\infty)\to[0,\infty]$ defined by

$\displaystyle\mu_{f}(\nu)=\mu\left(\{t\in\Omega:|f(t)|>\nu\}\right).$

It is easy to see that distribution function $\mu_{f}$ depends only on $|f|$ and is non-increasing. The decreasing rearrangement of $f$ is denoted by $f^{*}:[0,\infty)\to[0,\infty]$ and is defined as

$\displaystyle f^{*}(z)=\inf\{\nu\geqslant 0:\mu_{f}(\nu)\leqslant z\},\>z\in[0% ,\infty)$

where, by convention, $inf\{\emptyset\}=\infty$ . It is be noted that for any $\displaystyle\nu\geqslant 0,∼{}\mu\left(L(|f|,\nu)\right)=m\left(L(f^{*},\nu)% \right),$ here $m$ is the Lebesgue measure on $\mathbb{R}$ .

It is well known that the distribution function of $f$ is a right continuous function (see e.g. Bennett et al., 1988; Kristiansson, 2002). However, it need not be left continuous. Furthermore, if $m$ is the Lebesgue measure on $\mathbb{R}$ , then we have the following relation between $f^{*}$ and $\mu_{f}$ (see Kristiansson, 2002)

$\displaystyle f^{*}(z)=m(\{\nu\geqslant 0:\mu_{f}(\nu)>z\})=m_{\mu_{f}}(z),\>z% \geqslant 0.$ (1)

The above relation shows that $f^{*}$ is itself a distribution function with respect to the Lebesgue measure $m$ on the interval $[0,\infty)$ and is therefore a right continuous function. For more details and properties of these functions, reader can refer to (Bennett et al., 1988; Duff, 1967, Hardy et al., 1964). We say that the set $\{t\in\Omega:f(t)=a\}$ is a level set of $f$ at $a$ , for $a\in\mathbb{R}.$ Under the assumption that $\mu_{f}(0)<\infty$ , we prove that the distribution function is continuous at $\nu>0$ if only if the measure of the level set of $|f|$ at $\nu$ is zero. Using the same we derive a similar result for the continuity of the function $f^{*}$ .

2. Continuity of distribution function

In this section we prove the main result, which provides a necessary and sufficient condition for the distribution function of a given measurable function to be continuous. For the same we use the following theorem which ensures the continuity of Lebesgue measure.

.

Fitzpatrick et al., 2010 (Continuity of measure from above) Let $(\Omega,\Sigma,\mu)$ be a measure space and $\left\{S_{k}\right\}_{k=1}^{\infty}$ be a descending sequence of measurable sets with $\mu(S_{1})<\infty$ , then

$\mu\left(\bigcap\limits_{k=1}^{\infty}S_{k}\right)=\displaystyle\lim_{k\to% \infty}\mu\left(S_{k}\right).$

.

Let $(\Omega,\Sigma,\mu)$ be a measure space and $f:\Omega\to\mathbb{R}$ be a measurable function. Suppose $\nu>0$ and $\displaystyle\mu\left(\{t\in\Omega:|f(t)|>0\}\right)<\infty$ . Then the distribution function $\mu_{f}$ is continuous at $\nu$ if and only if $\mu(\{t\in\Omega:|f(t)|=\nu\})=0$ .

Proof..

Let $\nu>0$ be given. By Archimedian property of real numbers there exists $N\in\mathbb{N}$ satisfying $\nu>\frac{1}{N}$ . Therefore we have

$\displaystyle\nu-\frac{1}{n}>0\quad\forall∼{}n\geqslant N.$ (2)

For $i\geqslant N$ , set $A_{i}^{\nu}=\left\{t\in\Omega:|f(t)|>\nu-\frac{1}{i}\right\}.$ Then

$A_{N}^{\nu}\subseteq A_{N+1}^{\nu}\subseteq A_{N+2}^{\nu}\subseteq\ldots$

From Eq. (2) and monotonicity of measure, it follows that

$\displaystyle\mu(A_{N}^{\nu})=\mu\left(\left\{t\in\Omega:|f(t)|>\nu-\frac{1}{N% }\right\}\right)\leqslant\mu(\{t\in\Omega:|f(t)|>0\})<\infty.$

Therefore by continuity of measure (Theorem 1), we have

$\displaystyle\mu\left(\bigcap\limits_{n=N}^{\infty}A_{n}^{\nu}\right)=% \underset{n\rightarrow\infty}{\lim}\>\mu\left(A_{n}^{\nu}\right).$ (3)

Now

$\displaystyle\mu\left(\bigcap\limits_{n=N}^{\infty}A_{n}^{\nu}\right)=\mu\left% (\bigcap\limits_{n=N}^{\infty}\left\{t\in\Omega:|f(x)|>\nu-\frac{1}{n}\right\}% \right)=\mu\left(\{t\in\Omega:|f(t)|\geqslant\nu\}\right)=\mu(\{t\in\Omega:|f(% t)|>\nu\}\cup\{t\in\Omega:|f(t)|=\nu\})=\mu(\{t\in\Omega:|f(t)|>\nu\})+\mu(\{t% \in\Omega:|f(t)|=\nu\}).$ (4)

Thus from Eqs (3) and (4) we have,

$\displaystyle\lim_{n\to\infty}\mu\left(A_{n}^{\nu}\right)=\lim_{n\to\infty}\mu% \left(\left\{t\in\Omega:|f(t)|>\nu-\frac{1}{n}\right\}\right)=\mu(\{t\in\Omega% :|f(t)|>\nu\})+\mu(\{t\in\Omega:|f(t)|=\nu\}).$

Therefore

$\displaystyle\lim_{n\to\infty}\mu_{f}\left(\nu-\frac{1}{n}\right)=\mu_{f}(\nu)% +\mu(\{t\in\Omega:|f(t)|=\nu\}).$ (5)

Now assume that $\mu_{f}$ is continuous at $\nu$ then it is also left continuous at $\nu$ , so

$\displaystyle\lim_{n\to\infty}\mu_{f}\left(\nu-\frac{1}{n}\right)=\mu_{f}(\nu).$

Hence by Eq. (5)

$\displaystyle\mu(\{t\in\Omega:|f(t)|=\nu\})=0.$

Conversely, let $\mu(\{t\in\Omega:|f(t)|=\nu\})=0$ , then again by Eq. (5),

$\displaystyle\lim_{n\to\infty}\mu_{f}\left(\nu-\frac{1}{n}\right)=\mu_{f}(\nu).$

Thus $\mu_{f}$ is left continuous and hence continuous at $\nu$ . ∎

The following corollary is an immediate consequence of the above theorem.

Corollary 1. If $f$ is any strictly monotone function on $\Omega$ with $\mu_{f}(0)<\infty$ , then its distribution function is continuous.

The next example illustrates the Theorem 2.

Example 1. Let $C$ be the Cantor set and $\phi$ be the Cantor function on $[0,1]$ . Suppose $I^{1}_{1}=\displaystyle\left(\frac{1}{3},\frac{2}{3}\right)$ and for each $k>1$ , $I^{k}_{j}∼{}(1\leqslant j\leqslant 2^{k-1}$ ) are the removed intervals during the construction of the Cantor set $C$ . It is well known that the Cantor function is strictly increasing on $C$ and it stays constant on intervals $I^{k}_{j}$ for each $k\in\mathbb{N}∼{}(1\leqslant j\leqslant 2^{k-1}).$ The constant values are

$\displaystyle\phi(t)=\frac{2j-1}{2^{k}},\>t\in I_{j}^{k}∼{}(1\leqslant j% \leqslant 2^{k-1},\>k\in\mathbb{N}).$

Thus, for any $\nu=\displaystyle\frac{2j-1}{2^{k}}$ for some $k\in\mathbb{N},1\leqslant j\leqslant 2^{k-1}$ , we have

$\displaystyle\mu(\{t\in[0,1]:|\phi(t)|=\nu\})=\mu(I^{k}_{j})=\frac{1}{3^{k}}>0.$

Also, if $\nu\notin\displaystyle\left\{\frac{2j-1}{2^{k}}:1\leqslant j\leqslant 2^{k-1},% k\in\mathbb{N}\right\}$ , then it is easy to see that $\{t\in[0,1]:|\phi(t)|=\nu\}$ is either an empty or a singleton set. Hence $\displaystyle\mu\left(\{t\in[0,1]:|\phi(t)|=\nu\}\right)=0$ . Therefore by Theorem 2, the set of discontinuities of $\mu_{\phi}$ is $\displaystyle\left\{\frac{2j-1}{2^{k}}:1\leqslant j\leqslant 2^{k-1},k\in% \mathbb{N}\right\}.$

3. Continuity of decreasing rearrangement

It is obvious from the definition of Lebesgue measure on $\mathbb{R}$ that, if $A\subset\mathbb{R}$ is either an empty or a singleton set, then $m(A)=0$ , where $m$ is the Lebesgue measure on $\mathbb{R}$ . The following lemma ensures that the converse of the same is true for any level set of the distribution function.

Lemma 1. Let $(\Omega,\Sigma,\mu)$ be a measure space and $f:\Omega\to\mathbb{R}$ be a measurable function. The set $\{\nu\geqslant 0:\mu_{f}(\nu)=z\},\>z\geqslant 0$ is either empty or singleton if and only if its Lebesgue measure is zero.

Proof..

It is sufficient to prove that the set, $\{\nu\geqslant 0:\mu_{f}(\nu)=z\},\>z\geqslant 0$ is either empty or singleton if its Lebesgue measure is zero. Suppose not, then there exists $z\geqslant 0$ and two real numbers $0\leqslant\nu_{1}<\nu_{2}$ such that $\nu_{i}\in\{\nu\geqslant 0:\mu_{f}(\nu)=z\}$ for $i=1,2$ and $m(\{\nu\geqslant 0:\mu_{f}(\nu)=z\})=0.$ Here $m$ is the Lebesgue measure on $\mathbb{R}$ . Therefore we have,

$\displaystyle\mu_{f}(\nu_{1})=z=\mu_{f}(\nu_{2}).$ (6)

For any $\nu^{{}^{\prime}}\in[\nu_{1},\nu_{2}]$ , as $\mu_{f}$ is non-increasing, we have

$\displaystyle\mu_{f}(\nu_{2})\leqslant\mu_{f}(\nu^{{}^{\prime}})\leqslant\mu_{% f}(\nu_{1})$ (7)

Hence by Eqs (6) and (7), $\mu_{f}(\nu^{{}^{\prime}})=z\ \forall\>\nu^{{}^{\prime}}\in[\nu_{1},\nu_{2}].$ Therefore,

$\displaystyle m(\{\nu\geqslant 0:\mu_{f}(\nu)=z\})\geqslant m([\nu_{1},\nu_{2}% ])=\nu_{2}-\nu_{1}>0.$

A contradiction, this completes the proof. ∎

Now we use Theorem 2, to establish a necessary and sufficient condition for the decreasing rearrangement of a function to be continuous at some point.

.

Let $(\Omega,\Sigma,\mu)$ be a measure space and $f:\Omega\to\mathbb{R}$ be an $L^{\infty}$ function. Then the decreasing rearrangement $f^{*}$ of $f$ is continuous at $z>0$ if and only if $\{\nu\geqslant 0:\mu_{f}(\nu)=z\}$ is either empty or singleton.

Proof..

By Eq. (1), we have

$\displaystyle f^{*}(z)=m_{\mu_{f}}(z)=m(\{\nu\geqslant 0:\mu_{f}(\nu)>z\}),\>z% \geqslant 0.$

Therefore

$\displaystyle m(\{\nu\geqslant 0:\mu_{f}(\nu)>0\})=f^{*}(0)=\inf\{\nu\geqslant 0% :\mu_{f}(\nu)\leqslant 0\}=\inf\{\nu\geqslant 0:\mu(\{t\in\Omega:|f(t)|>\nu\})% =0\}\quad(\because\>\mu_{f}(\nu)\geqslant 0)=\underset{\Omega}{\textit{ess sup% }}∼{}|f(t)|<\infty.$ (8)

By Eq. (8), Theorem 2 and Lemma 1, it follows that $f^{*}$ is continuous at $z>0$ if and only if the set $\{\nu\geqslant 0:\mu_{f}(\nu)=z\}$ , is either empty or singleton. ∎

The following example illustrates the above theorem.

Example 2. Let $f:[0,2]\to\mathbb{R}$ be defined by

$\displaystyle f(t)=\left\{\begin{array}[]{ll}t^{2}&\text{if}\>0\leqslant t% \leqslant 1\\ 1&\text{if}\>1<t\leqslant 2.\end{array}\right.$

Then its distribution function is given by

$\displaystyle\mu_{f}(\nu)=\left\{\begin{array}[]{ll}2-\sqrt{\nu}&\text{if}\>0% \leqslant\nu<1\\ \par 0&\text{if}\>\nu\geqslant 1.\end{array}\right.$

Clearly for every positive real number $z$ the set $\{\nu\geqslant 0:\mu_{f}(\nu)=z\}$ is either empty or singleton. Therefore, by Theorem 3, $f^{*}$ is continuous at each positive real number. Since $f^{*}$ is a right continuous function, so it is continuous at zero as well. It follows that, $f^{*}$ is a continuous function. In fact, $f^{*}$ can be calculated be

$\displaystyle f^{*}(z)=\left\{\begin{array}[]{ll}1&\text{if}\>0\leqslant z% \leqslant 1\\ (2-z)^{2}&\text{if}\>1<z<2\\ 0&\text{if}\>z\geqslant 2.\end{array}\right.$

4. Conclusion

The decreasing rearrangement $f^{*}$ , of a given measurable function $f$ , is calculated by a two step process, the preceding one is to determine the distribution function $\mu_{f}$ and then to obtain $f^{*}$ . For an arbitrary measurable function, obtaining $\mu_{f}$ and/or $f^{*}$ is sometimes a daunting task. The results in this paper provide methods to determine the points of continuity (or discontinuity) of $\mu_{f}$ and $f^{*}$ , without explicit form of these functions. More precisely, the continuity of the distribution function $\mu_{f}$ is characterized by the measure of the level sets of $|f|$ , whereas the continuity of $f^{*}$ is characterized by the number elements of the level sets of $\mu_{f}$ .

Footnotes

Acknowledgments

The first author would like to thank UGC for financial support.

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