# ANALYSIS II: integration

1. Apr 21, 2008

### jjou

Let $$A$$ be a rectangle in $$\mathbb{R}^k$$, $$B$$ be a rectangle in $$\mathbb{R}^n$$, $$Q=AxB$$. If $$f:Q\rightarrow\mathbb{R}$$ is a bounded function and $$\int_Q f$$ exists, show that $$\int_B f(x,y)$$ exists for every $$x\in A-D$$ where $$D$$ is a set of measure zero in $$\mathbb{R}^n$$.

I know that, for fixed $$x\in A$$, $$\int_B f(x,y)$$ exists if $$D_x=\{y\in B|f\mbox{ is discontinuous at }(x,y)\}$$ has measure zero.

A set has measure zero if, for any $$\epsilon>0$$, there exists a covering of that set by countably many rectangles $$D_1, D_2, ...$$ such that $$\sum_i v(D_i)<\epsilon$$ (where $$v(D_i)$$ is the volume of $$D_i$$).

My idea was to let $$D$$ be the set of $$x\in A$$ such that $$D_x$$ does not have measure zero. Then, clearly, we have that $$\int_B f(x,y)$$ exists for $$x\in A-D$$. Then I would only have to show that $$D$$ has measure zero. I would do this by showing that $$f$$ is not continuous at points in $$D$$ (obvious). Since $$f$$ is integrable over $$Q$$, $$D$$ must have measure zero in $$Q$$.

My problem is that I know that a set $$D$$ which does not have measure zero in $$A$$ might have measure zero in $$Q$$ (since $$Q$$ is in a higher dimension). Is this not the right approach?