Folland: Why is the product measure well-defined?

Question

Consider the following fragment from Folland's "Real analysis" on p64:

I don't understand why $\pi(E)$ is well-defined, i.e. assume that $$E = \bigcup_i A_i \times B_i= \bigcup_j C_j \times D_j$$ where the unions are disjoint. Then why is it true that $$\sum_i \mu(A_i)\nu(B_i) = \sum_j \mu(C_j)\nu(D_j)?$$

Folland talks about a common refinement, but I don't see how that works.

Answer 1

Note that we can write $E=\bigcup_{i,j}(A_i\times B_i)\cap(C_j\times D_j)=\bigcup_{i,j}(A_i\cap C_j)\times(B_i\cap D_j)$.

So: $A_i\times B_i=\bigcup_{j}(A_i\cap C_j)\times(B_i\cap D_j)$, $C_j\times D_j=\bigcup_{i}(A_i\cap C_j)\times(B_i\cap D_j)$.

So we have: $\mu(E)=\mu(\bigcup_i A_i\times B_i)=\mu(\bigcup_{i,j}(A_i\cap C_j)\times(B_i\cap D_j))=\sum_{i,j}\mu_1(A_i\cap C_j)\mu_2(B_i\cap D_j)$

Now:

$\sum_{i}\mu_1(A_i)\mu_2(B_i)=\sum_{i,j}\mu_1(A_i\cap C_j)\mu_2(B_i\cap D_j)=\sum_j\mu_1(C_j)\mu_2(D_j)$, where the last step is true because I can change order of summation (everything is non-negative).

Answer 2

Let's show how a common refinement works in a concrete example. This will get the idea across, and it's fairly clear that this works for any region you like.

Say $E$ is a set that can be written as a disjoint union of finitely many rectangles. For instance, this region (pardon the paint):

Now we define the area (measure) of this region to be the sum of the areas (measures) of the rectangles in a decomposition. For instance, we can witness this region as a disjoint union

and define the area of $E$ to be the sum of the areas of these three rectangles (of course, we know what the area of a rectangle should be).

But wait, you say! What if instead we chose this decomposition:

then the area of $E$ should be the sum of the areas of these three rectangles! But that might give us a different answer. How can we guarantee that we get the same result no matter which decomposition we choose?

Here's the secret: Overlay the two pictures on top of each other:

When we do this, it subdivides each of our previous regions into more rectangles.

When we subdivide our rectangles, we call this a refinement of our covering. But we've cleverly chosen these new rectangles to be a refinement of both of our coverings of interest. That is, it's a common refinement.

And now we see that the area of $E$ as measured by the first covering is the same as the area measured by the second covering. Why? Because both are the same as the area measured by the purple covering. After all,

$$ \begin{aligned} \text{area}(\text{red}_1 + \text{red}_2 + \text{red}_3) &= \text{area}( \text{purple}_1 + (\text{purple}_2 + \text{purple}_3 + \text{purple}_4) + (\text{purple}_5 + \text{purple}_6 ) \\ &= \text{area}( \text{purple}_2 + (\text{purple}_3 + \text{purple}_5) + (\text{purple}_1 + \text{purple}_4 + \text{purple}_6) ) \\ &= \text{area}( \text{blue}_1 + \text{blue}_2 + \text{blue}_3) \end{aligned} $$

Where I haven't labelled the red/blue regions explicitly (mainly because I'm lazy), but hopefully it's clear from context.

So we see that, as long as we can find a common refinement, we can run this same argument. The area as computed by a covering is the same as the area computed by any of its refinements (this is basically additivity of a measure on disjoint sets), so if two coverings have a common refinement, they must give the same area.

Now, since a common refinement always exists (this is annoying to prove formally, but should be intuitively obvious), this means any two coverings of the same region give the same area, which is exactly what we wanted to show!

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I hope this helps ^_^