Separating a double integral when f depends only on x and g on y

[carlosbgois]

Hi there. I think I have proved on little theorem on double integrals, showed below. Are my arguments 'correct' (I mean, rigorous enough)?

Let $f$ be a function of $x, f(x)$, $g$ be a function depending only on $y, g(y)$, and last, let $A$ be the set determined by $a≤x≤b$ and $c≤y≤d$. By Fubini's theorem, $\int\int_{A}f(x)g(y)dxdy=\int^{d}_{c}[\int^{b}_{a}f(x)g(y)dx]dy.$ Knowing $g$ is a function only dependent on $y$, it is a constant when integrating over x, hence we have $\int^{d}_{c}[g(y)\int^{b}_{a}f(x)dx]dy.$ Still, as $f$ depends only on $x$, so does any integral over $f$, then finally $\int^{d}_{c}g(y)dy \int^{b}_{a}f(x)dx.$

Many thanks.

 

[DonAntonio]

Hi there. I think I have proved on little theorem on double integrals, showed below. Are my arguments 'correct' (I mean, rigorous enough)?

Let $f$ be a function of $x, f(x)$, $g$ be a function depending only on $y, g(y)$, and last, let $A$ be the set determined by $a≤x≤b$ and $c≤y≤d$. By Fubini's theorem, $\int\int_{A}f(x)g(y)dxdy=\int^{d}_{c}[\int^{b}_{a}f(x)g(y)dx]dy.$ Knowing $g$ is a function only dependent on $y$, it is a constant when integrating over x, hence we have $\int^{d}_{c}[g(y)\int^{b}_{a}f(x)dx]dy.$ Still, as $f$ depends only on $x$, so does any integral over $f$, then finally $\int^{d}_{c}g(y)dy \int^{b}_{a}f(x)dx.$

Many thanks.

Kudos if you reached this result by yourself, but in fact it follows in a pretty simple way from well known properties of integrals.

DonAntonio

 

[carlosbgois]

Yes it does follows easily, nice to know it is correct. It can simplify many problems as I was testing just now. Thanks!