The Unapologetic Mathematician

Mathematics for the interested outsider

Adding and Multiplying Measurable Real-Valued Functions

One approach to the problem of adding and multiplying measurable functions on a measurable space X 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 (\mathbb{R}^2,\mathcal{B}_2) like we did before to treat the additive identity (0,0) specially. Then we could set up products (which we will eventually do) and get a map (f,g):X\to\mathbb{R}^2 and compose this with the Borel map (x,y)\mapsto x+y or the Borel map (x,y)\mapsto xy. 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 f and g are extended real-valued measurable functions on a measurable space (X,\mathcal{S}) and if c is any real number, then each of the sets

\displaystyle\begin{aligned}A&=\left\{x\in X\vert f(x)<g(x)+c\right\}\\B&=\left\{x\in X\vert f(x)\leq g(x)+c\right\}\\C&=\left\{x\in X\vert f(x)=g(x)+c\right\}\end{aligned}

has a measurable intersection with every measurable set. If X is itself measurable, of course, this just means that these three sets are measurable.

To see this for the set A, consider the (countable) set \mathbb{Q}\subseteq\mathbb{R} of rational numbers. If f(x) really is strictly less than g(x)+c, then there must be some rational number r between them. That is, if x\in A then for some r we have f(x)<r and r-c<g(x). And thus we can write A 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}

By the measurability of f and g, this is the countable union of a collection of measurable sets, and is thus measurable.

We can write B as X\setminus\left\{x\in X\vert f(x)<g(x)-c\right\}, and so the assertion for B follows from that for A. And we can write C=B\setminus A, 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, \left[fg\right](x)=\infty if and only if f(x)=g(x)=\pm\infty. Since the sets f^{-1}(\{\infty\})\cap g^{-1}(\{\infty\}) and f^{-1}(\{-\infty\})\cap g^{-1}(\{-\infty\}) are both measurable, the set [fg]^{-1}(\{\infty\}) — their union — is measurable. We can handle [fg]^{-1}(\{-\infty\}), [f+g]^{-1}(\{\infty\}), and [f+g]^{-1}(\{-\infty\}) similarly.

So now we turn to our convenient condition for measurability. Since we’ve handled the sets where f(x) and g(x) are infinite, we can assume that they’re finite. Given a real number c, we find

\displaystyle\left\{x\in X\vert f(x)+g(x)<c\right\}=\left\{x\in X\vert f(x)<c-g(x)\right\}

which is measurable by our lemma above (with -g in place of g). Since this is true for every real number c, the sum f+g is measurable.

To verify our assertion for the product fg, 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

\displaystyle f(x)g(x)=\frac{1}{4}\left(\left(f(x)+g(x)\right)^2-\left(f(x)-g(x)\right)^2\right)

We just found that the sum f+g and the difference f-g 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 fg is a scalar multiple of the difference of these squares, and is thus measurable.

May 7, 2010 - Posted by | Analysis, Measure Theory


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  5. Minor mistake: you missed a g^{-1} on the second line of re-writing A.

    Comment by Son | February 17, 2013 | Reply

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