## 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

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, if and only if . Since the sets and are both measurable, the set — their union — is measurable. We can handle , , and 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.

[…] and Negative Parts of Functions Now that we have sums and products to work with, we find that the maximum of and — sometimes written or — and their […]

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[…] and some limit inferior . If these two coincide, then the sequence has a proper limit . But one of our lemmas tells us that the set of points where any two measurable functions coincide has a nice property: […]

Pingback by Sequences of Measurable Functions « The Unapologetic Mathematician | May 10, 2010 |

[…] function is simple — . And thus the collection of simple functions forms a subalgebra of the algebra of measurable […]

Pingback by Simple and Elementary Functions « The Unapologetic Mathematician | May 11, 2010 |

[…] finally we can show that converges in measure to . We can use the same polarization trick as we’ve used before. Write ; we’ve just verified that the squares converge to squares, and we know that linear […]

Pingback by Convergence in Measure and Algebra « The Unapologetic Mathematician | May 21, 2010 |

Minor mistake: you missed a on the second line of re-writing .

Comment by Son | February 17, 2013 |