IMT Exercise 1.1.4

Let {d_1,d_2\ge1}, and let {E_1\subset\mathbb{R}^{d_1}}, {E_2\subset\mathbb{R}^{d_2}} be elementary sets. Show that {E_1\times E_2\subset\mathbb{R}^{d_1+d_2}} is elementary , and {m^{d_1+d_2}(E_1\times E_2)=m^{d_1}(E_1)\times m^{d_2}(E_2)}.
Let partitions of {E_1} and {E_2} into finite sets of disjoint boxes be {B^1_1\cup\ldots\cup B^1_n}, and {B^2_1\cup\ldots\cup B^2_m}. Then

\displaystyle E_1\times E_2=\bigcup_{\substack{i=1\ldots n\\ j=1\ldots m}} B^1_i\times B^2_j

Consider {B^1_i=I^1_{i,1}\times\ldots\times I^1_{i,d_1}} and {B^2_j= I^2_{j,1}\times\ldots\times I^2_{j,d_2}}, then
\displaystyle B^1_i\times B^2_j=I^1_{i,1}\times\ldots\times I^1_{i,d_1}\times I^2_{j,1}\times\ldots\times I^2_{j,d_2}

It is clear that {B^1_i\times B^2_j\subset\mathbb{R}^{d_1+d_2}} and as {d_1} and {d_2} are both finite so therefore is {d_1+d_2}. Furthermore
\displaystyle m^{d_1+d_2}(B^1_i\times B^2_j)=|I^1_{i,1}|\times\ldots\times |I^1_{i,d_1}|\times |I^2_{j,1}|\times\ldots\times |I^2_{j,d_2}|=m^{d_1}(B^1_i)\times m^{d_2}(B^2_j).

As {E_1} and {E_2} are partitioned into disjoint boxes the new boxes in the Cartesian product {E_1\times E_2} are also disjoint. This gives
\displaystyle m^{d_1+d_2}(E_1\times E_2)=\sum_{\substack{i=1\ldots n\\ j=1\ldots m}} m^{d_1}(B^1_i)\times m^{d_2}(B^2_j)=m^{d_1}(E_1)\times m^{d_2}(E_2)

{\Box}

Advertisements
This entry was posted in IMT Exercises, Measure Theory. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s