Fubini's Theorem: Is This True?

[Tatianaoo]

Does anybody know if the following is true?

Let $p>1$. If $f=f(x,y)$ is such that $f\in L^p([a,b]\times [a,b])$, then $f_y(x)\in L^p([a,b])$ for almost all $y\in[a,b]$ and $f_x(y)\in L^p([a,b])$ for almost all $x\in[a,b]$.

Is this a consequence of Fubini's theorem?

 

[Vargo]

Unfortunately, $ don't work. You need to enclose your code with [/itex] and [itex]. The one with the slash goes at the end but I had to put them in reverse order for it to show up (or it would have just treated "and" as my code.

 

[Vargo]

The answer to your question is yes. g is Lp if and only if |g|^p is L1. So apply Fubini to the function $|f(x,y)|^p$, which is L^1.

 

[Tatianaoo]

Thanks a lot for help!

 

[blue_raver22]

Yes, this statement is a consequence of Fubini's theorem. Fubini's theorem states that if $f$ is a measurable function on a product space $X\times Y$, where $X$ and $Y$ are measurable spaces, and if $f$ is integrable over $X\times Y$ with respect to the product measure, then the iterated integrals of $f$ over $X$ and $Y$ are also integrable and equal to the integral of $f$ over $X\times Y$. In this case, $f$ is integrable over $[a,b]\times [a,b]$ with respect to the product measure, and therefore, the iterated integrals $f_y(x)$ and $f_x(y)$ are also integrable over $[a,b]$ and equal to the integral of $f$ over $[a,b]\times [a,b]$. This implies that $f_y(x)$ and $f_x(y)$ are in the $L^p$ space for almost all $y\in[a,b]$ and $x\in[a,b]$, respectively. Therefore, the statement in question is true.