Switching order of integration on unbounded domain

Question

Let's say ${f\left( {x,y} \right)}$ is a continuous function and assume that:

$\int\limits_{ - \infty }^\infty {\left| {f\left( {x,y} \right)} \right|dx} $ converges for all $y $

2) $\int\limits_{ - \infty }^\infty {\left| {f\left( {x,y} \right)} \right|dy} $ converges for all $x $

Is the following statement correct? $$\int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {f\left( {x,y} \right)dxdy} } = \int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {f\left( {x,y} \right)dydx} } $$

This is supposed to be some version of Fubini's or Tonelli's theorem but I wasn't able to find the exact version for this case.

Answer 1

I think the conclusion is false.

Let $g\colon \mathbb{R}\to\mathbb{R}$ be a continuous non-negative function, such that $g(x) = 0$ if $x\not\in [0,1]$, and $\int_0^1 g = 1$. Let us consider the function $$ f(x,y) = \sum_{k=0}^\infty [g(x-k) - g(x-k-1)] g(y-k), \qquad (x,y)\in\mathbb{R}^2. $$ Observe that, for every $(x,y)\in\mathbb{R}^2$, at most one term of the series is different from $0$, and that assumptions 1) and 2) are satisfied.

Since, if $x\in [n, n+1]$, $n\in\mathbb{N}$, one has $$ f(x,y) = \begin{cases} g(x)g(y), & \text{if}\ n = 0, \\ g(x-n)[g(y-n) - g(y-n-1)], & \text{if}\ n\geq 1, \end{cases} $$ it is easy to see that $$ \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} f(x,y)\, dy = 1. $$ On the other hand, if $y\in[n,n+1]$, $$ f(x,y) = [g(x-n) - g(x-n-1)] g(y-n), $$ hence $\int_{-\infty}^{+\infty} f(x,y) dx = 0$ for every $y\in\mathbb{R}$, and $\int_{-\infty}^{+\infty} dy \int_{-\infty}^{+\infty} f(x,y) dx = 0$ as well.

Answer 2

There's something of a standard library of analysis counterexamples. Looking here... iterated integrals don't match, but the inner integrals are unconditional... OK, I've got it. Well, the version I have in memory is for sums: $\begin{bmatrix}1&0&0&0&\cdots\\-1&1&0&0&\cdots\\0&-1&1&0&\cdots\\0&0&-1&1&\cdots\\ \vdots&\vdots&\vdots&\vdots&\ddots\end{bmatrix}$

Sum along the rows first (inner) and we get $1$. Sum along the columns first and we get zero.

This can of course be converted into a version for the integral. Let $$f(x,y)=\begin{cases}1& 0<x-y<1\\ -1&-1<x-y<0\\0&\text{otherwise}\end{cases}$$ for $x,y\ge 0$. Then \begin{align*}\int_0^{\infty}f(x,y)\,dx &= \begin{cases}1-y& 0\le y\le 1\\ 0&y>1\end{cases}\\ \int_0^{\infty}f(x,y)\,dy &= \begin{cases}x-1& 0\le x\le 1\\ 0&x>1\end{cases}\end{align*} so $\int_0^{\infty}\int_0^{\infty}f(x,y)\,dx\,dy = \frac12$ and $\int_0^{\infty}\int_0^{\infty}f(x,y)\,dy\,dx = -\frac12$. The order of the integral cannot be interchanged in general.

Now, quibbles. That's not continuous! No problem; just smooth the jumps out when defining $f$ - say, one period of a sawtooth wave $f(x,y)=g(x-y)$ where $g(t)=\begin{cases}t&-\frac12\le t\le \frac12\\1-t&\frac12\le t\le 1\\-1-t&-1\le t\le -\frac12\\ 0&\text{otherwise}\end{cases}$. Calculating the integral is a little messier, but we'll still get something positive for $\int_0^{\infty} f(x,y)\,dx$ when $y\in (0,1)$ and something negative for $\int_0^{\infty} f(x,y)\,dy$ when $x\in (0,1)$.
That's an integral on $[0,\infty)^2$, not $\mathbb{R}^2$! Padding the function with zeros when either variable is negative would do the trick - except that it disrupts continuity again. So then, we need to tweak it so that it's zero on the axes. OK - multiply $f$ by $\min\{x,y,1\}$. Working this out for my sawtooth example, it's a lot of calculation for not much benefit, so I'll just give the answer: $$\int_0^{\infty} f(x,y)\,dx=\begin{cases}0& y\ge 2\\ \frac{(2-y)^3}{6}& \frac32\le y\le 2\\ \frac{y(\frac32-y)^2}{6}+\frac{5-3y}{24}& 1\le y\le \frac32\\ \frac y4 - \frac{3y-1}{24} - \frac{(2-y)(y-\frac12)^2}{6} & \frac12 \le y\le 1\\ \frac y4-\frac{y^3}{6}&0\le y\le \frac12\\ 0&\text{otherwise}\end{cases}$$ All but one of those are clearly nonnegative; we simplify that third case to clear the last bit up: \begin{align*}\frac y4 - \frac{3y-1}{24} - \frac{(2-y)(y-\frac12)^2}{6} &= \frac{4y-3y+1-4y^3+12y^2-9y+2}{24}\\ &= \frac{-4y^3 + 12y^2 - 8y + 1}{24}\\ &= \frac1{24}-\frac{y(1-y)(2-y)}{6}\\ \frac y4 - \frac{3y-1}{24} - \frac{(2-y)(y-\frac12)^2}{6} &\ge \frac1{24}-\frac{(1-y)(2-y)}{6}\ge 0\end{align*} (The inequalities, of course, use that $\frac12 \le y\le 1$) So then, $\int_0^{\infty} f(x,y)\,dx$ is always nonnegative. By the symmetry of $f$'s definition, $\int_0^{\infty} f(x,y)\,dy$ is always nonpositive, and the two iterated integrals are unequal.

This example runs on essentially the same principles as Rigel's already posted answer, but I find it a little clearer to work with the explicit piecewise definition than the fake infinite sum.