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Let‎ ‎$‎‎f:‎\mathbb{R}\to‎\mathbb{R}‎$ be a probability density function. Can ‎the following be happened for ‎$‎‎f$?

(1) ‎‎$‎‎f$ ‎is ‎not ‎integrable ‎on ‎an ‎(some) ‎interval ‎of ‎‎$\mathbb{R}‎$.

‎(2) ‎‎$‎‎f$ ‎is ‎not ‎integrable ‎on every closed ‎interval ‎of ‎‎$\mathbb{R}‎$.‎‎‎ ‎ I know that ‎if $‎f‎$ ‎is a‎ ‎‎probability density function then

(1) ‎$‎‎f(x)‎\geq‎0 ‎\quad‎\text{for all} \; x$,

(2) ‎$\int_{-‎\infty‎}^{+\infty}f(x)\,dx=1$.

but ‎here ‎we ‎have‎ Lebesgue integral not Riemann integral. Moreover if ‎$‎‎f$ ‎wants ‎to ‎be‎ Riemann integrable on the whole $\mathbb{R}‎$, it must hold in the following conditions

‎‎(a) ‎‎$‎‎f$ ‎is ‎integrable ‎on every closed ‎interval ‎of ‎$\mathbb{R},‎$

(b) the following integral is convergent‎

$$\int_{-‎\infty‎}^{+\infty}f(x)\,dx=‎‎\int_{-‎\infty‎}^{0}f(x)\,dx+\int_{0‎}^{+\infty}f(x)\,dx.$$

According the mentioned things, the most pdf are ‎ Riemann integrable, and I could not find any example as I asked. Would anyone help me to find that. thanks a lot. ‎

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  • $\begingroup$ Take $X\sim U[0,1]$. Then $g=1-1_{\mathbb{Q}}$ is a version of $f_X$... $\endgroup$
    – user140541
    Commented Dec 20, 2018 at 7:37

2 Answers 2

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It is known that there exists a measurable set $E$ in $\mathbb R$ such that $0<m(E\cap I) <m(I)$ for every open interval $I$. If $f=\frac {I_E} {m(E)}$ then $f$ is a density function but it is not continuous at any point so it is not Riemann integrable on any interval.

For the construction of such a set $E$ see Creating a Lebesgue measurable set with peculiar property.

It is easy give simpler examples where $f$ is almost everywhere equal to a Riemann integrable function but it is not itself Riemann integarble. In probability theory density function which are equal almost everywhere lead to the same distribution, so I tried to give a example which is not almost everywhere equal to a Riemann integrable function.

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  • $\begingroup$ This is very nice since, also, $f$ is not a.e. equal to a Riemann integrable function. $\endgroup$ Commented Dec 20, 2018 at 7:44
  • $\begingroup$ This is a very interesting example. Is the existence of such a set $E$ easy to prove? $\endgroup$
    – BigbearZzz
    Commented Dec 20, 2018 at 8:27
  • $\begingroup$ I have added a reference for that construction. @BigbearZzz $\endgroup$ Commented Dec 20, 2018 at 8:51
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$$ f(x) = \begin{cases} 0 &; x\in\Bbb Q \cap[0,1] \\ 1 &; x\in\Bbb Q^c \cap[0,1] \end{cases} $$ is a probability density function that is not Riemann integrable.

If you want a density function that is not Riemann integrable on any interval in $\Bbb R$ then you can take $f$ to be the (normalized) Gaussian distribution and then $f+\mathbf 1_{\Bbb Q}$ is the desired probability density function with the properties you want.

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