# Measurable Functions behaviour on $\mathbb{R}^{2}$

A measurable rectangle in the cartesian product $X \times Y$ of two measurable spaces in $\mathbb{R}^{2}$ is of the form $A \times B$, where $A$ and $B$ are measurable subsets of $X$ and $Y$ respectively.

Question Can a measurable function on $\mathbb{R}^{2}$ be essentially unbounded on every measurable rectangle of positive measure?

What I did was to recognize a measurable function as limit of sequence of bounded measurable functions, but don't really know how to proceed from there.

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What is the source of the problem? –  Bill Dubuque Aug 14 '10 at 21:50
@Jason De Vito: How can we prove it for the Conway base function. –  anonymous Aug 14 '10 at 22:05
@Jason: The Conway Base 13 function is zero almost everywhere, so it is essentially bounded. –  Pete L. Clark Aug 14 '10 at 22:48
@Chandru1: Your question and its setup are copied verbatim from Halmos. Would you please give attribution in your question? –  Jonas Meyer Oct 18 '10 at 23:58

## 3 Answers

The answer to this is yes! Let $g\colon\mathbb{R}\to\mathbb{R}$ be essentially unbounded on every nontrivial open interval and set $f(x,y)=g(x+y)$. Then, f is essentially unbounded on every measurable rectangle $A\times B$ of positive Lebesgue measure.

To construct g you can take $g_0(x)=1_{\{1>x>0\}}x^{-1/2}$ which is integrable but not square integrable on any neighborhood of 0. Then, letting $q_1,q_2,\ldots$ be an enumeration of the rationals, $g(x)=\sum_n2^{-n}g_0(x-q_n)$ is integrable (so, is finite almost everywhere) but not square integrable on any open interval, so is not essentially bounded.

To see that this construction of $f$ works, pick any two sets $A,B \subseteq \mathbb{R}$ with positive Lebesgue measure. By intersecting with a bounded interval, we can restrict to sets with finite measure. I'll write $I_A(x)$ for the indicator function of a set A and $(I_A*I_B)(x)=\int I_A(x-y)I_B(y)\,dy$ for the convolution, which is a continuous function. For any $K>0$ let $C=\{x\in\mathbb{R}\colon\vert g(x)\vert\ge K\}$, which has positive measure in any nontrivial open interval. Then, \begin{align} \mu((A\times B)\cap \{x\in\mathbb{R}^2\colon \vert f(x)\vert\ge K\})&=\int\int I_A(x)I_B(y)I_C(x+y)\,dx\,dy\\ &=\int (I_A*I_B)(x)I_C(x)\,dx. \end{align} The convolution $I_A*I_B$ is a nonnegative function with positive integral $\int (I_A*I_B)(x)\,dx=\mu(A)\mu(B)>0$, so it must be strictly positive at some point. By continuity, it is bounded below by a positive number on some open interval, so the integral displayed above is positive. Therefore, $f$ is essentially unbounded on $A\times B$.

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I'll add that this idea is a refinement of the proof that A+B contains an open interval for sets A,B of positive measure, using the fact that $I_A*I_B$ is continuous and not identically zero. –  George Lowther Aug 19 '10 at 1:21
Ah, yes, the function you construct is a common counterexample (In PDE), that I didn't think of that myself :). Nice! It looks correct to me. It is a different approach than mine, so I'm curious if mine works too. –  Jonas Teuwen Aug 19 '10 at 9:04

Maybe the following works:

Let $R_n$ be a countable basis of rectangles (for example rational coordinates) for $\mathbb{R}^2$. Define $f_0 = 0$ and given $f_n$ define $f_{n + 1}$ as $f_n$ outside $R_{n + 1}$ and $n + 1$ inside $R_{n + 1}$. $(f_n)$ is an increasing sequence of measurable functions so $f = \lim f_n$ is measurable.

Now take any open rectangle $R$ and given $M > 0$ take $n > M$ such that $R_n \subset R$. Now $\mu(|f| > M) \geq \mu(|f_n| > M) \geq \textrm{area}(R_n) > 0$. So it is essentially unbounded.

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Please refer Paul Halmos's book entitled "Problems for Mathematicians Young and Old". Its somewhere located in the Measure theory Section. Although, i couldn't comprehend the solution fully, i would like to see a better example.

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What do you call a "better" example? –  Jonas Teuwen Aug 19 '10 at 20:08