Does there exist any probability density function ‎$‎f:‎\mathbb{R}\to‎\mathbb{R}‎$ ‎which is not Riemann integrable? 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.
‎

$$
f(x) = \begin{cases}
0 &; x\in\Bbb Q \cap[0,1] \\
1 &; x\in\Bbb Q^c \cap[0,1]
\end{cases}
$$
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.
