Suppose we have a continuous distribution function $F$ with density $f$ such that $f$ is Lebesgue integrable. Does $F$ then also have a density that is Riemann integrable?
Here's an attempt to construct a counterexample. Take the density $$ f(x)= \begin{cases} 1 & \text{if } x \in [0,1]\setminus\mathbb{Q},\\ 0 & \text{otherwise}. \end{cases} $$ The Lebesgue integral of this function is $1,$ yet it isn't Riemann integrable. However, one could just use $$ \tilde{f}(x)= \begin{cases} 1 & \text{if } x \in [0,1],\\ 0 & \text{otherwise}. \end{cases} $$ instead which has the same cumulative distribution function, but is also Riemann integrable.
So my question is, can such a modification always be carried out?

  • $\begingroup$ I am not sure if this is what you need as a counter-example. The Cantor function: en.wikipedia.org/wiki/Cantor_function $\endgroup$ – herb steinberg May 3 '18 at 0:08
  • $\begingroup$ One way is to consider an everywhere unbounded function on $[0,1]$, which you can construct like $f(x) =\sum_n a_n(x-q_n) ^{1/2}$, where the $q_n$ is an enumeration of the rationals in $[0,1]$ and $a_n$ an appropriate sequence of positive numbers. $\endgroup$ – Jose27 May 3 '18 at 0:19
  • $\begingroup$ @herbsteinberg That's a distribution which doesn't have any density at all - Lebesgue or Riemann integrable. But I'm already assuming that I have existence and Lebesgue integrability. $\endgroup$ – Sebastian Oberhoff May 3 '18 at 0:20
  • $\begingroup$ The probability distribution for which your example is a density is the same as that for which the indicator of the unit interval is a density, i.e. it is the uniform distribution. $\endgroup$ – Michael Hardy May 3 '18 at 1:42
  • $\begingroup$ @herbsteinberg : The Cantor distribution is not an example of the kind requested here, since it doesn't have a density at all. $\endgroup$ – Michael Hardy May 3 '18 at 1:43

Consider a set $C$ like the standard Cantor set, but of measure $>0$. Take the characteristic function of it. Note that $C$ closed and the interior of $C$ is empty. Moreover, $C$ is such that for every open interval $I$ that intersects $C$ we have $\mu(C\cap I)>0$. Therefore, for every $x \in C$ and $I$ open interval around $x$ we have $\mu(C\cap I)>0$ and $\mu((\mathbb{R}\setminus C)\cap I)>0$.

Consider a function $f$ that is equal almost everywhere to $\chi_C$. We conclude from the above that for every $x\in C$ and every $I$ open interval around $x$, $f$ takes both values $0$ and $1$ in $I$. Therefore $f$ cannot be continuous at $x$. So the set of discontinuities of $f$ contains $C$, and so is of measure $>0$. Therefore, $f$ is not Riemann integrable.

Conclusion: there does not exist a Riemann integrable function $f$ such that $$\int_0^x f(t) dt =\int_0^x \chi_C(x) dx $$ for all $x$.

Added: This should not be confused with the Cantor distribution, which is a continuous, but singular distribution.

The Cantor distribution is like this : take the standard Cantor set $C$ (of measure $0$) There exists a Borel measure $\nu$ on $C$ that has total mass $1$. ( since $C$ is like $\{0,1\}^{\mathbb{N}}$) . Define the distribution on $\mathbb{R}$ by $m([0,x])=\nu(C\cap [0,x])$. This gives a continuous, but singular distribution on $\mathbb{R}$, supported on the set $C$, of measure $0$.

  • $\begingroup$ Note that the following look different from each other: $$ \begin{align} & \mathbb C \setminus C \\ \\ & \mathbb C \backslash C \end{align} $$ The former uses \setminus and the latter uses \backslash. The latter doesn't have the spacing appropriate to a binary operation symbol. $\endgroup$ – Michael Hardy May 3 '18 at 1:46
  • $\begingroup$ What is "a distribution of function type"? $\qquad$ $\endgroup$ – Michael Hardy May 3 '18 at 1:48
  • $\begingroup$ Just showing $f$ is not Riemann-integrable is not enough. The function given in the question is also not Riemann-integrable, but the probability distribution for which it is a density does have a Riemann-integrable density, which is almost everywhere equal to the given function. You need to show that $f$ is not almost everywhere equal to any Riemann-integrable function. $\endgroup$ – Michael Hardy May 3 '18 at 1:51
  • $\begingroup$ @Michael Hardy; what is shown is that any function that is equal to $\chi_C$ almost everywhere will be discontinuous at every point of $C$. So there is no Riemann integrable function that is equal to $\chi_C$ almost everywhere. Hope that makes it clearer. $\endgroup$ – Orest Bucicovschi May 3 '18 at 2:02
  • $\begingroup$ Could you add something explicit about that to the answer? $\endgroup$ – Michael Hardy May 3 '18 at 2:12

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