It is well-known that one can get the Lebesgue measure on [0, 1] by tossing a fair coin infinitely (countably) many times and mapping each sequence to a real number written out in binary.

I was trying to explore what happens if you follow the same procedure with a biased coin. I managed to prove that if the induced measure is absolutely continuous with respect to the Lebesgue measure, then the density must be discontinuous on at least a dense set (which includes all rational numbers in [0, 1] with a finite binary expansion).

But I suspect it's much worse than that. When I try to visualize the cumulative distribution function, it seems to have a "fractal staircase" type of shape, so I suspect the measure is, in fact, singular, but I can't prove it.

Q1. Are there any standard techniques for proving a measure is singular ?

Q2. I strongly suspect that a natural question like this has already been resolved. Any link to a textbook and/or paper would be appreciated.

  • $\begingroup$ Reference: P. Billingsley "Probability and Measure" Third Edition. page 407. (From comments to a MSE question.) $\endgroup$ – r.e.s. Sep 20 '16 at 14:24

For $\omega \in [0,1]$, let $X_i(\omega)$ be the $i$th binary digit of $\omega$. (If $\omega$ is a dyadic rational and thus has two binary expansions, let's say we choose the expansion that ends with all 0s; it makes no difference). Under Lebesgue measure, the $X_i$ are iid Bernoulli $1/2$ random variables, so by the strong law of large numbers, $\frac{1}{n}(X_1 + \dots + X_n) \to 1/2$ almost surely. In other words, the set $A_{1/2} := \{\omega : \lim_{n \to \infty} \frac{1}{n} (X_1(\omega) + \dots + X_n(\omega)) = \frac{1}{2}\}$ has Lebesgue measure $1$.

If $\mu_p$ is the measure coming from a coin that comes up heads with probability $p \ne 1/2$, then under $\mu_p$, the $X_i$ are iid Bernoulli $p$. By the strong law again, we have $\mu_p(A_p) = 1$. But $A_p$ and $A_{1/2}$ are clearly disjoint, so Lebesgue measure and $\mu_p$ are mutually singular.

I guess the "standard technique" here is to use a zero-one type theorem from probability giving an almost sure statement about limiting behavior, and look for a place where the two measures exhibit different limiting behavior.

  • $\begingroup$ This is a nice answer - I think it's essentially the same as mine, but yours has the advantage of relying on more well-known technology (the SLLN). $\endgroup$ – Anthony Quas Sep 20 '16 at 8:50
  • $\begingroup$ When you have a hammer, everything looks like a nail. I agree that a more elementary argument (such as Nate's) is better. $\endgroup$ – Gerald Edgar Sep 20 '16 at 17:28
  • $\begingroup$ This is a truly excellent answer. I am keeping this as Exhibit A for how a probabilistic argument neatly proves a result which seems very intractable by analysis type techniques. I suspect the "hard analysis" part goes into proving the strong law of large numbers. $\endgroup$ – Anindya Sep 20 '16 at 23:23
  • $\begingroup$ @Anindya: Sort of, but proving the strong law in this case isn't particularly difficult. $\endgroup$ – Nate Eldredge Sep 20 '16 at 23:59
  • $\begingroup$ This classical argument can give more, namely the Hausdorff dimension of the measures $\mu_p$ and the sets $A_p$. For an exposition and references see, e.g., math.stonybrook.edu/~bishop/Bishop_Peres_Jan_22_2015.pdf $\endgroup$ – Yuval Peres Sep 23 '16 at 17:32

Yes. You're right; the measure is singular. One soft way to do this is to use ergodic theory. The "coin-tossing measures" are all distinct ergodic shift-invariant measures on $\{0,1\}^{\mathbb N}$ and any two ergodic invariant measures for any fixed transformation are mutually singular. The map from sequences of 0's and 1's to $[0,1]$ is 1-1 off a countable set, and so mutual singularity is preserved when the measures are transferred to $[0,1]$.

One reference for all of this is Peter Walters' book, An Introduction to Ergodic Theory (publ. Springer Graduate Texts in Mathematics).

  • 1
    $\begingroup$ I swtiched to Nate's answer as this was a very direct argument which nails the result. I will follow up on your Ergodic Theory reference though. It's very interesting that a pretty simple and natural question is leading into quite deep waters. Thanks. $\endgroup$ – Anindya Sep 20 '16 at 23:26
  • $\begingroup$ I'm glad you got what you wanted. The Birkhoff ergodic theorem that is at the heart of my answer is a generalization of the Strong Law of Large Numbers, so the answers are, in fact, quite similar even though they look different. $\endgroup$ – Anthony Quas Sep 20 '16 at 23:32

Just a side comment: if you pass from binary to trinary, you can still obtain the Lebesgue measure by choosing the digits $0$, $1$, $2$, with probabilities $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ (something one may call "a fair trinary coin"). Now, if you use the "biased" coin $(\frac{1}{2},0,\frac{1}{2})$, then you'll arrive exactly at the Cantor function, which puts mass $1$ to the Cantor set (which has null Lebesgue measure).

  • 1
    $\begingroup$ That's really interesting. Great observation ! $\endgroup$ – Anindya Sep 20 '16 at 23:30

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