## Composing Real-Valued Measurable Functions II

As promised, today we come up with an example of a measurable function and a Lebesgue measurable function so that the composition is not measurable. Specifically, will be the closed unit interval , considered as a measurable subspace of .

Now, every point can be written out in ternary as

We set (depending on ) to be the first index for which , and if no such index exists. Then we define the function

That is, write out the number in ternary until you hit a , and leave off everything after that. Change all the s to s, and consider the resulting string of s and s as a number written out in binary. The extra fraction added in the formula above comes from that first . This function is often called the “Cantor function” because of its relationship to the famous Cantor set. In case it’s not apparent, the Cantor set is the collection of points with no .

First of all, is increasing from to . Clearly , so ; and so . Given points and , if then for and . If and or , then as we write out in binary the th bit is , while the th bit of is and so . On the other hand, if and , then the th bit of both and is , but stops at that point while has at least one more bit equal to . And so again .

Maybe more surprising is the fact that is actually continuous! If again we have and and for , then we find

Thus, given an we can find a large enough so that . Then we can pick a small enough so that two numbers differing by less than will agree to the first places in their ternary expansions, and so is continuous.

Unfortunately, might not be *strictly* increasing. Indeed, on any stretch of , the function is actually constant! It’s interesting to note that manages to increase continuously from to while remaining constant almost everywhere. But still we’re going to need a *strictly* increasing function for our purposes. We get this by considering . This still increases continuously from to , but now it’s strictly increasing.

But as a strictly increasing continuous function from to itself, it has a strictly increasing continuous inverse. That is, there is a strictly increasing continuous function such that if and only if . And since it’s continuous, it’s Borel measurable, and any Borel measurable function is Lebesgue measurable.

Now, the set is Lebesgue measurable and has positive measure. This is the collection of points of the form for . To get at this, first we consider the collection . It’s pretty straightforward to see that this consists of all terminating binary expansions, which are exactly the rational numbers. But this is a countable set, and countable sets have Lebesgue measure zero. Consequently, we find that . Since , there must be some positive measure in in order to make up the difference.

But now we can take a thick, non-Lebesgue measurable set whose intersection with is itself a non-Lebesgue measurable set . However, , and has Lebesgue measure zero. Since every subset of a set of Lebesgue measure zero is itself Lebesgue measurable (by completeness), must be Lebesgue measurable, even though is not. This is not a problem because we only ever asked that the preimage under of any *Borel* set be Lebesgue measurable.

At last, we set — the characteristic function of this set . This function is Lebesgue measurable, because the preimage of any set is one of , , or , all of which are Lebesgue measurable. And we’ve already established that is measurable. However, the composition is *not* measurable, since the preimage of the Borel set is

which is not Lebesgue measurable.

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