Csdrhrt

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

$intlimits_{ - infty }^infty {left| {fleft( {x,y} right)} right|dx} $ converges for all $y $

2) $intlimits_{ - infty }^infty {left| {fleft( {x,y} right)} right|dy} $ converges for all $x $

Is the following statement correct?
$$intlimits_{ - infty }^infty {intlimits_{ - infty }^infty {fleft( {x,y} right)dxdy} } = intlimits_{ - infty }^infty {intlimits_{ - infty }^infty {fleft( {x,y} right)dydx} } $$

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.

I think the conclusion is false.

Let $gcolon mathbb{R}tomathbb{R}$ be a continuous non-negative function, such that
$g(x) = 0$ if $xnotin [0,1]$, and $int_0^1 g = 1$.
Let us consider the function
$$
f(x,y) = sum_{k=0}^infty [g(x-k) - g(x-k-1)] g(y-k),
qquad (x,y)inmathbb{R}^2.
$$
Observe that, for every $(x,y)inmathbb{R}^2$, at most one term of the series is different from $0$, and that assumptions 1) and 2) are satisfied.

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

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: $begin{bmatrix}1&0&0&0&cdots\-1&1&0&0&cdots\0&-1&1&0&cdots\0&0&-1&1&cdots\ vdots&vdots&vdots&vdots&ddotsend{bmatrix}$

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 $x,yge 0$. Then
begin{align*}int_0^{infty}f(x,y),dx &= begin{cases}1-y& 0le yle 1\ 0&y>1end{cases}\
int_0^{infty}f(x,y),dy &= begin{cases}x-1& 0le xle 1\ 0&x>1end{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 $g(t)=begin{cases}t&-frac12le tle frac12\1-t&frac12le tle 1\-1-t&-1le tle -frac12\ 0&text{otherwise}end{cases}$. Calculating the integral is a little messier, but we'll still get something positive for $int_0^{infty} f(x,y),dx$ when $yin (0,1)$ and something negative for $int_0^{infty} f(x,y),dy$ when $xin (0,1)$.

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:
$$int_0^{infty} f(x,y),dx=begin{cases}0& yge 2\ frac{(2-y)^3}{6}& frac32le yle 2\ frac{y(frac32-y)^2}{6}+frac{5-3y}{24}& 1le yle frac32\ frac y4 - frac{3y-1}{24} - frac{(2-y)(y-frac12)^2}{6} & frac12 le yle 1\ frac y4-frac{y^3}{6}&0le yle frac12\ 0&text{otherwise}end{cases}$$
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 0end{align*}
(The inequalities, of course, use that $frac12 le yle 1$)
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.