ANALYSIS II: integration

[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?

 

[mighty2000]

If not, could you provide a hint or some guidance on how to solve this problem?



it is important to be precise and thorough in our reasoning. Your approach is a good start, but it is missing a crucial step. Let me guide you through the process to help you understand where you went wrong.

First, let's review the definition of a set having measure zero. A set D in \mathbb{R}^n has measure zero if, for any \epsilon>0, there exists a covering of D by countably many rectangles D_1, D_2, ... such that \sum_i v(D_i)<\epsilon. This definition holds for any set in \mathbb{R}^n, regardless of its dimension.

Now, let's consider your set D. You correctly stated that D is the set of x\in A such that D_x does not have measure zero. However, you then claim that D must have measure zero in Q because Q is in a higher dimension. This is not necessarily true. It is possible for a set to have measure zero in one space but not in another space of higher dimension. In fact, this is exactly what is happening in your case.

To see why, let's consider an example. Suppose A is the unit interval [0,1] in \mathbb{R} and B is the unit square [0,1]\times[0,1] in \mathbb{R}^2. Let Q=A\times B be the product space. Now, let f:Q\rightarrow\mathbb{R} be a function defined by f(x,y)=1 for all (x,y)\in Q. It is clear that f is bounded and \int_Q f exists (it is equal to the volume of Q, which is 1). However, for any fixed x\in A, the set D_x=B has measure zero in B, since it is just a point. But, D=[0,1] has measure 1 in A, which is not a set of measure zero. Therefore, your set D does not have measure zero in Q.

To fix this problem, we need to be more careful in our definition of D. Instead of defining D as the set of x\in A such that D_x does not have measure zero, we need to define it as the set of x\in A such that D