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Let $\mu$ be a finite measure on $\mathbb R$ which has no atoms, and no component continuous with respect to Lebesgue measure. An example is the law of the random variable $$ \sum_{k\ge 1}3^{-k}X_k $$ where the $X_k$ are IID Bernoulli (0-1 valued) random variables, because $\mu$ only charges numbers which ternary expansion contains only $0$s and $1$s.

Is it always possible to find $n\ge 1$ such that the $n-$th convolution product of $\mu$, denoted by $\mu^{\otimes n}$, has a non-zero component with respect to Lebesgue measure? With the example above, the answer is yes with $n=2$ because if one takes $X_n'$ independent copies of the $X_n$, then $\mu^{\otimes 2}$ is the law of $$ \sum_{k\ge 1}3^{-k}(X_k+X_k'), $$ which charges all ternary expansions.

On the other hand, my experience with ugly measures is that the answer should be no, but I can't find a counter-example...

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    $\begingroup$ Your example could be something like $\sum X_k/k!$, with $X_k$ IID Bernoulli. $\endgroup$
    – Gerald Edgar
    Commented Apr 30, 2021 at 9:13
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    $\begingroup$ Is this different from the the question "Self convolutions of singular continuous mesure"? mathoverflow.net/questions/266450/… $\endgroup$
    – user95282
    Commented Apr 30, 2021 at 12:00

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I don't believe that you can a absolutely continuous measure convolving finitely many cantors. The fourier transform for the cantor measure is $ \hat f = \Pi cos(\frac {2 \pi t} {3^n}) $ and it has the feature that $\hat f(3^n) = \hat f (1) $, esp., it does not go to zero, and therefore nor does any power, showing that no power is absolutely continuous. $$$$ This does not rule out that it might have an absolutely continuous component. If you want to dispose of that possibility, Gerald's example will $$$$

  1. be singular non atomic, and $$$$
  2. have the property that the lim sup of $\hat f$ is 1, which will also be true of any power, showing that the measure is singular.
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