Adding and Multiplying Measurable Real-Valued Functions
One approach to the problem of adding and multiplying measurable functions on a measurable space 
would be to define a two-dimensional version of Borel sets and Lebesgue measure, and to tweak the definition of a measurable function to this space 
like we did before to treat the additive identity 
specially. Then we could set up products (which we will eventually do) and get a map 
and compose this with the Borel map 
or the Borel map 
. In fact, if you’re up for it, you can go ahead and try working out this approach as an exercise.
Instead, we’ll take more of a low road towards showing that the sum and product of two measurable functions are measurable. We start with a useful lemma: if 
and 
are extended real-valued measurable functions on a measurable space 
and if 
is any real number, then each of the sets
has a measurable intersection with every measurable set. If is itself measurable, of course, this just means that these three sets are measurable.
To see this for the set 
, consider the (countable) set 
of rational numbers. If 
really is strictly less than 
, then there must be some rational number 
between them. That is, if 
then for some we have 
and 
. And thus we can write as the countable union
![\displaystyle\begin{aligned}A&=\bigcup\limits_{r\in\mathbb{Q}}\left(\left\{x\in X\vert f(x)<r\right\}\cap\left\{x\in X\vert r-c<g(x)\right\}\right)\\&=\bigcup\limits_{r\in\mathbb{Q}}\left(f^{-1}\left[-\infty,r\right]\cap\left[r-c,\infty\right]\right)\end{aligned}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7DA%26%3D%5Cbigcup%5Climits_%7Br%5Cin%5Cmathbb%7BQ%7D%7D%5Cleft%28%5Cleft%5C%7Bx%5Cin+X%5Cvert+f%28x%29%3Cr%5Cright%5C%7D%5Ccap%5Cleft%5C%7Bx%5Cin+X%5Cvert+r-c%3Cg%28x%29%5Cright%5C%7D%5Cright%29%5C%5C%26%3D%5Cbigcup%5Climits_%7Br%5Cin%5Cmathbb%7BQ%7D%7D%5Cleft%28f%5E%7B-1%7D%5Cleft%5B-%5Cinfty%2Cr%5Cright%5D%5Ccap%5Cleft%5Br-c%2C%5Cinfty%5Cright%5D%5Cright%29%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}A&=\bigcup\limits_{r\in\mathbb{Q}}\left(\left\{x\in X\vert f(x)<r\right\}\cap\left\{x\in X\vert r-c<g(x)\right\}\right)\\&=\bigcup\limits_{r\in\mathbb{Q}}\left(f^{-1}\left[-\infty,r\right]\cap\left[r-c,\infty\right]\right)\end{aligned}")
By the measurability of and , this is the countable union of a collection of measurable sets, and is thus measurable.
We can write 
as 
, and so the assertion for follows from that for . And we can write 
, so the statement is true for that set as well.
Anyway, now we can verify that the sum and product of two measurable extended real-valued functions are measurable as well. We first handle infinite values separately. For the product, =\infty](https://s0.wp.com/latex.php?latex=%5Cleft%5Bfg%5Cright%5D%28x%29%3D%5Cinfty&bg=e6e6e6&fg=333333&s=0 "\left[fg\right](x)=\infty")
if and only if 
. Since the sets 
and 
are both measurable, the set ![[fg]^{-1}(\{\infty\})](https://s0.wp.com/latex.php?latex=%5Bfg%5D%5E%7B-1%7D%28%5C%7B%5Cinfty%5C%7D%29&bg=e6e6e6&fg=333333&s=0 "[fg]^{-1}(\{\infty\})")
— their union — is measurable. We can handle ![[fg]^{-1}(\{-\infty\})](https://s0.wp.com/latex.php?latex=%5Bfg%5D%5E%7B-1%7D%28%5C%7B-%5Cinfty%5C%7D%29&bg=e6e6e6&fg=333333&s=0 "[fg]^{-1}(\{-\infty\})")
, ![[f+g]^{-1}(\{\infty\})](https://s0.wp.com/latex.php?latex=%5Bf%2Bg%5D%5E%7B-1%7D%28%5C%7B%5Cinfty%5C%7D%29&bg=e6e6e6&fg=333333&s=0 "[f+g]^{-1}(\{\infty\})")
, and ![[f+g]^{-1}(\{-\infty\})](https://s0.wp.com/latex.php?latex=%5Bf%2Bg%5D%5E%7B-1%7D%28%5C%7B-%5Cinfty%5C%7D%29&bg=e6e6e6&fg=333333&s=0 "[f+g]^{-1}(\{-\infty\})")
similarly.
So now we turn to our convenient condition for measurability. Since we’ve handled the sets where and 
are infinite, we can assume that they’re finite. Given a real number , we find
which is measurable by our lemma above (with 
in place of ). Since this is true for every real number , the sum 
is measurable.
To verify our assertion for the product 
, we turn and recall the polarization identities from when we worked with inner products. Remember, they told us that if we know how to calculate squares, we can calculate products. Something similar is true now, as we write
We just found that the sum and the difference 
are measurable. And any positive integral power of a measurable function is measurable, so the squares of the sum and difference functions are measurable. And then the product is a scalar multiple of the difference of these squares, and is thus measurable.
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Minor mistake: you missed a
on the second line of re-writing 
.