Riemann sums for a Lebesgue integrable density function Let $f$ be a non-negative measurable function whose support is $[0,1]$ and it's integrable. In other words, it's a density function of a continuous random variable which is $[0,1]$-valued. Then is it true that for any fixed $z \in [0,1]$, we have
$$
\displaystyle \sum_{m=0}^{n-1} \frac{1}{n} f\left( \frac{m}{n}+\frac{z}{n}\right) \xrightarrow{n \rightarrow \infty} \int_{0}^1 f(x) \ dx
$$
Does it hold only when $f$ is continuous or does it hold in general for any such density function ?

No. The indicator function of the rationals satisfies $\int_0^1\mathbf 1_\Bbb Q(x)\,dx=0$ but for $z=0$, $\sum_{m=0}^{n-1}\frac1n 1_\Bbb Q(m/n)=1\not\to0$ as $n\to\infty$.
It should be clear from this example that you can make these two diverge from each other however you like, since the sum samples only countably many points, which have no bearing on the value of the integral. Of course if $f$ is continuous then the theorem is true, but you seem to already know that.
