A counter-example to the reduction theorem of integrals

[Leonardo891]

For [tex]$ i \in \mathbb{N}^+ $[/tex] let [tex]$ \phi_i \in C( \mathbb{R} ) $[/tex] have support in [tex]$ (2^{-i},2^{1-i}) $[/tex] , such that [tex]$ \int \phi_i = 1 $[/tex] . Put

[tex]$ f(x,y)= \sum_{i=1}^\infty [ \phi_i(x) - \phi_{i+1} (x) ] \phi_{i} (y) $[/tex], [tex]$ f: \mathbb{R}^2 \to \mathbb{R}$[/tex], [tex]$ f=0 $[/tex] where not specified.

Then [tex]$ f $[/tex] has compact support in [tex]$ \mathbb{R}^2 $[/tex], [tex]$ f $[/tex] is continuous except at [tex]$ (0,0) $[/tex], and

[tex]$ \int dy \int f(x,y) dx = 0 $[/tex] but [tex]$ \int dx \int f(x,y) dy = 1 $[/tex]

Observe that [tex]$ f $[/tex] is unbounded in every neighborhood of [tex]$ (0,0) $[/tex]
The integrals are in the sense of Peano-Jordan (the normal sense).

I have a my solution (I hope correct ) and I will post it next Sunday.
Good work!

[Leonardo891]

"Rigel":The support of a function $f$ is usually defined as the closure of the set $\{f\ne 0\}$, so that it is closed by definition.

Ok, I'm an idiot...
I (didn't solve) solved this part trying to prove the sequential compactness of the support and I actually proved that given a sequence in the support, it's possible to find a Cauchy's subsequence but I forgot that the completeness of the support is necessary. Obviously a closed subspace of a complete space is complete but if I had remembered that supports are closed...

Actually this problem was easier than that I thought.

[Rigel1]

"Leonardo89":Ok for the continuity but...
[quote="Rigel"]$"spt" f$ is compact, since it is bounded (as already shown) and closed (by definition).

The boundness is obvious but I don't understand because the support is closed by definition (I solved this part in another way, maybe more complicated )[/quote]
The support of a function $f$ is usually defined as the closure of the set $\{f\ne 0\}$, so that it is closed by definition.

[Leonardo891]

Ok for the continuity but...

"Rigel":$"spt" f$ is compact, since it is bounded (as already shown) and closed (by definition).

The boundness is obvious but I don't understand because the support is closed by definition (I solved this part in another way, maybe more complicated ).

[Rigel1]

$"spt" f$ is compact, since it is bounded (as already shown) and closed (by definition).
Moreover, for every open set $A\subset\mathbb{R}^2$ s.t. $bar{A}\subset\mathbb{R}^2\setminus\{(0,0)\}$, there exists a finite set $I_A \subset\mathbb{N}^+$ of indices s.t. $f(x,y) = \sum_{i\in I_A}[\phi_i(x) - \phi_{i+1}(x)]\phi_i(y)$ for every $(x,y)\in A$. Hence $f$ is continuous in $A$ for every such $A$, so that it is continuous in $\mathbb{R}^2\setminus\{(0,0)\}$.

[Leonardo891]

Rigel your part of solution is exactly identical to mine. Now it remain to prove the continuity of [tex]$f$[/tex] except that in [tex]$(0,0)$[/tex] (nearly obvious) and the compactness of the support.

[Rigel1]

From the very definition of $f$ we have that $"spt" f \subset (0,1)\times (0,1)$.

For every fixed $y\in (0,1)$ there exists a unique $k\in\mathbb{N}^+$ s.t. $y\in [2^{-k}, 2^{1-k})$; moreover $f(x,y) = [\phi_k(x)-\phi_{k+1}(x)]\phi_k(y)$, and
$\int f(x,y) dx = \phi_k(y) \int [\phi_k(x)-\phi_{k+1}(x)] dx = 0$.
We conclude that $\int dy \int f(x,y) dx = 0$.

On the other hand, let us fix $x\in (0,1)$.
If $x\in [1/2, 1)$, then $f(x,y) = \phi_1(x) \phi_1(y)$, and $\int f(x,y) dy = \phi_1(x)$.
If $x\in [2^{-k}, 2^[1-k})$, $k\ge 2$, then $f(x,y) = \phi_k(x) [\phi_k(y) - \phi_{k-1}(y)]$, and $\int f(x,y) dy = 0$.
Hence we conclude that $\int dx \int f(x,y) dy = \int_{1/2}^{1} \phi_1(x) dx = 1$.