The Unapologetic Mathematician

Mathematics for the interested outsider

Composing Real-Valued Measurable Functions II

As promised, today we come up with an example of a measurable function f:X\to\mathbb{R} and a Lebesgue measurable function \phi:\mathbb{R}\to\mathbb{R} so that the composition (\phi\circ f):X\to\mathbb{R} is not measurable. Specifically, (X,\mathcal{S}) will be the closed unit interval \left[0,1\right], considered as a measurable subspace of (\mathbb{R},\mathcal{L}).

Now, every point x\in\left[0,1\right] can be written out in ternary as

\displaystyle x=\sum\limits_{i=1}^\infty\frac{\alpha_i}{3^i}=.\alpha_1\alpha_2\alpha_3\dots

We set n (depending on x) to be the first index for which \alpha_n=1, and n=\infty if no such index exists. Then we define the function

\displaystyle\psi(x)\sum\limits_{1\leq i<n}\frac{\alpha_i}{2^{i+1}}+\frac{1}{2^n}

That is, write out the number in ternary until you hit a 1, and leave off everything after that. Change all the 2s to 1s, and consider the resulting string of 0s and 1s as a number written out in binary. The extra fraction added in the formula above comes from that first \alpha_n=1. 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 C of points with no \alpha_n=1.

First of all, \psi is increasing from 0 to 1. Clearly 0=.000\dots, so \psi(0)=0; and 1=.222\dots so \psi(1)=1. Given points x=.\alpha_1\alpha_2\alpha_3\dots and y=.\beta_1\beta_2\beta_3\dots, if x<y then \alpha_i=\beta_i for 1\leq i<j and \alpha_j<\beta_j. If \alpha_j=0 and \beta_j=1 or \beta_j=2, then as we write out \psi(y) in binary the jth bit is 1, while the jth bit of \psi(x) is 0 and so \psi(x)<\psi(y). On the other hand, if \alpha_j=1 and \beta_j=2, then the jth bit of both \psi(x) and \psi(y) is 1, but \psi(x) stops at that point while \psi(y) has at least one more bit equal to 1. And so again \psi(x)<\psi(y).

Maybe more surprising is the fact that \psi is actually continuous! If again we have x=.\alpha_1\alpha_2\alpha_3\dots and y=.\beta_1\beta_2\beta_3\dots and \alpha_i=\beta_i for 1\leq i<j, then we find

\displaystyle\left\lvert\psi(y)-\psi(x)\right\rvert\leq\frac{1}{2^{j-1}}

Thus, given an \epsilon>0 we can find a large enough j so that \frac{1}{2^{j-1}}<\epsilon. Then we can pick a small enough \delta so that two numbers differing by less than \delta will agree to the first j places in their ternary expansions, and so \psi is continuous.

Unfortunately, \psi might not be strictly increasing. Indeed, on any stretch of x\in X\setminus C, the function \psi is actually constant! It’s interesting to note that \psi manages to increase continuously from 0 to 1 while remaining constant almost everywhere. But still we’re going to need a strictly increasing function for our purposes. We get this by considering y\mapsto\frac{1}{2}(y+\psi(y)). This still increases continuously from 0 to 1, but now it’s strictly increasing.

But as a strictly increasing continuous function from [0,1] to itself, it has a strictly increasing continuous inverse. That is, there is a strictly increasing continuous function f such that y=f(x) if and only if x=\frac{1}{2}(y+\psi(y)). And since it’s continuous, it’s Borel measurable, and any Borel measurable function is Lebesgue measurable.

Now, the set f^{-1}(C) is Lebesgue measurable and has positive measure. This is the collection of points of the form \frac{1}{2}(y+\psi(y)) for y\in C. To get at this, first we consider the collection \psi(X-C). 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 \mu(f^{-1}(X-C))=\frac{1}{2}. Since \mu(f^{-1}(X))=1, there must be some positive measure in f^{-1}(C) in order to make up the difference.

But now we can take a thick, non-Lebesgue measurable set whose intersection with f^{-1}(C) is itself a non-Lebesgue measurable set S. However, f(S)=M\subseteq C, and C has Lebesgue measure zero. Since every subset of a set of Lebesgue measure zero is itself Lebesgue measurable (by completeness), M must be Lebesgue measurable, even though f^{-1}(M)=S is not. This is not a problem because we only ever asked that the preimage under f of any Borel set be Lebesgue measurable.

At last, we set \phi=\chi_M — the characteristic function of this set M. This function \phi is Lebesgue measurable, because the preimage of any set is one of \emptyset, M, M^c or \mathbb{R}, all of which are Lebesgue measurable. And we’ve already established that f is measurable. However, the composition \phi\circ f is not measurable, since the preimage of the Borel set \{1\} is

\displaystyle(\phi\circ f)^{-1}(\{1\})=f^{-1}(\phi^{-1}(\{1\}))=f^{-1}(M)=S

which is not Lebesgue measurable.

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May 5, 2010 - Posted by | Analysis, Measure Theory

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