5
$\begingroup$

Question: Give an example of a Borel measurable function ${f: [0,1]^2 \rightarrow {\bf R}}$ such that the integrals ${\int_{[0,1]} f(x,y)\,\mathrm dy}$ and ${\int_{[0,1]} f(x,y)\,\mathrm dx}$ exist and are absolutely integrable for all ${x \in [0,1]}$ and ${y \in [0,1]}$ respectively, and that ${\int_{[0,1]} \left(\int_{[0,1]} f(x,y)\,\mathrm dy\right)\,\mathrm dx}$ and ${\int_{[0,1]} \left(\int_{[0,1]} f(x,y)\,\mathrm dx\right)\,\mathrm dy}$ exist and are absolutely integrable, but such that $$ \int_{[0,1]} \left(\int_{[0,1]} f(x,y)\,\mathrm dy\right)\,\mathrm dx \neq \int_{[0,1]} \left(\int_{[0,1]} f(x,y)\,\mathrm dx\right)\,\mathrm dy$$ are unequal. (Hint: adapt the doubly infinite sequence ${(x_{n,m})_{n,m \in {\bf N}}}$ such that ${x_{n,m}}$ equals ${+1}$ when ${n=m}$, ${-1}$ when ${n=m+1}$, and ${0}$ otherwise.) Thus we see that Fubini’s theorem fails when one drops the hypothesis that ${f}$ is absolutely integrable with respect to the product space.

Attempt: Partition the unit interval $[0,1]$ into $\left[\frac{1}{2^{k-1}}, \frac{1}{2^k}\right)$, $k=1, 2, \ldots$ as countably many sub-intervals of length $2^{-k}$. Denote by $I_n$ the $n^{\text{th}}$ interval in the $x$-axis, and $J_m$ the $m^{\text{th}}$ interval in the $y$-axis. Define $f:[0,1]^2 \rightarrow {\bf R}$ by $$f(x,y) := \sum_{n,m} x_{n,m} 1_{I_n \times J_m}(x,y).$$ It then seems that $$\int_{[0,1]} \left(\int_{[0,1]} f(x,y)\,\mathrm dy\right)\,\mathrm dx = \frac{1}{2} \neq 0 = \int_{[0,1]} \left(\int_{[0,1]} f(x,y)\,\mathrm dx\right)\,\mathrm dy.$$ But how about the part that emphasizes the absolute integrability of $f$ w.r.t. the product space?

$\endgroup$
3
  • 1
    $\begingroup$ I do not think your example gives a counterexample of the Fubini's Theorem as for your $f$, $|f|\le 1$ and therefore $\int_{[0,1]^2} |f|$ is finite, which satisfies the hypothesis of Fubini's Theorem. In order to adapt the example of sequence $(x_{n,m})$ in the hint, one should not set $f$ to be $x_{n,m}$ on $I_n\times J_m$. Instead one should construct $f$ so that its integration on $I_n\times J_m$ equals $x_{n,m}$. $\endgroup$ Commented yesterday
  • 3
    $\begingroup$ A minor remark: In Measure Theory, there is no such thing as an absolutely integrable function. Any integrable function is an absolutely integrable. $\endgroup$ Commented yesterday
  • $\begingroup$ @KaviRamaMurthy: All relevant definitions are from Terry Tao’s “An Introduction to Measure Theory” textbook. $\endgroup$ Commented yesterday

2 Answers 2

Reset to default
6
$\begingroup$

Take the famous example

$$f(x,y)=\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}$$

Then $\int_{0}^{1}f(x,y)\,dy=\frac{1}{1+x^{2}}$ (As $\frac{d}{dy}\frac{y}{x^{2}+y^{2}}=f(x,y)$)

which gives you $\int_{0}^{1}(\int_{0}^{1}f(x,y)dy)dx=\frac{\pi}{4}$.

On the other hand, $\int_{0}^{1}f(x,y)\,dx=-\frac{1}{1+y^{2}}$ and hence $\int_{0}^{1}(\int_{0}^{1}f(x,y)\,dx)\,dy)=-\frac{\pi}{4}$.

Where does this function fail Fubini's conditions?

It's because the function is not integrable (absolutely integrable) wrt the Lebesgue measure on $[0,1]^{2}$. To see this, restrict to the unit disc and use polar coordinates to say

\begin{align}\iint_{[0,1]^{2}}|\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}|&\,d(x,y)\geq\iint_{x^{2}+y^{2}\leq 1}|\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}|\,d(x,y)\\\\ &=\int_{0}^{2\pi}\int_{0}^{1}\frac{r^{2}|\cos^{2}(\theta)-\sin^{2}(\theta)|}{r^{4}}\cdot r\,dr\,d\theta=+\infty \end{align}

As also mentioned in the comments, for Lebesgue integrals, absolute integrability and integrability are the same things and Lebesgue integral is only defined for absolutely integrable functions.

$\endgroup$
2
  • $\begingroup$ Thanks, an edited solution is posted using the original hint of the problem. $\endgroup$ Commented 13 hours ago
  • $\begingroup$ @shark Yeah that's the standard infinite matrix example that works easily with the counting measure on $\Bbb{N}\times\Bbb{N}$. i.e. $x_{i,i}=1$, $x_{i,i+1}=-1$ for each $i$ and $x_{i,j}=0$ otherwise. Then the row sums are $0$ identically and thus the row-column sum is 0. But the column-row sum is $1$. $\endgroup$ Commented 6 hours ago
2
$\begingroup$

Let ${(x_{n,m})_{n,m \in {\bf N}}}$ be a doubly infinite sequence of extended non-negative reals such that ${x_{n,m}}$ equals ${+1}$ when ${n=m}$, ${-1}$ when ${n=m+1}$, and ${0}$ otherwise. Now, partition the unit interval $[0,1]$ into $[\frac{1}{2^{k-1}}, \frac{1}{2^k}),\ k=1, 2, \ldots$ as countably many sub-intervals of length $2^{-k}$. Denote by $I_n$ the $n^{th}$ interval in the $x$-axis, and $J_m$ the $m^{th}$ interval in the $y$-axis. Define $f:[0,1]^2 \rightarrow {\bf R}$ by \begin{align*} f(x,y) := \sum_{n,m} \frac{x_{n,m}}{|I_n| \cdot |J_m|} 1_{I_n \times J_m}(x,y). \end{align*} $\forall x \in [0,1],\ f(x,y) = \frac{1}{|I_n|} \sum_{m} \frac{x_{n,m}}{|J_m|} 1_{J_m}(y)$ for some fixed $n$. Specifically, \begin{align*} g(x) := \int_{[0,1]} f(x,y)\ dy = \begin{cases} 2, & x \in I_1 \\ 0, & x \in I_n\ \text{for}\ n \geq 2 \end{cases} \end{align*} Hence $g(x)$ is integrable, and we have \begin{align*} \int_{[0,1]} (\int_{[0,1]} f(x,y)\ dy)\ dx = \int_{[0,1]} g(x)\ dx = 1. \end{align*} $\forall y \in [0,1],\ f(x,y) = \frac{1}{|J_m|} \sum_{n} \frac{x_{n,m}}{|I_n|} 1_{I_n}(x)$ for some fixed $m$. Specifically, \begin{align*} h(y) := \int_{[0,1]} f(x,y)\ dx = 0. \end{align*} Hence $h(y)$ is integrable, and we have \begin{align*} \int_{[0,1]} (\int_{[0,1]} f(x,y)\ dx)\ dy = \int_{[0,1]} h(y)\ dy = 0. \end{align*} The function $f$ as defined is clearly Borel-measurable. But \begin{align*} |f(x,y)| = \sum_{n,m} \frac{1}{|I_n| \cdot |J_m|} 1_{I_n \times J_m}(x,y) \end{align*} is not integrable with respect to the product space.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.