Riemann like sums for Lebesgue integrable function Let $f(x)$ be a non-negative function on $\mathbb{R}$ such that $\int_{-\infty}^{\infty}f(x) \ dx=1.$ Actually, f(x) is a probability density function for a continuous random variable.
Can I justify that 
$$
\lim_{n \rightarrow \infty} \sum_{m \in\mathbb{Z}} \frac{1}{n} f(\frac{m}{n}+\frac{z}{n})= \int_{-\infty}^{\infty} f(x) \ dx
$$
where $z \in [0,1]$ is fixed. I am trying to use Lebesgue Dominated convergence Theorem but which function should I pick up as a dominator to justify the exchange of limits and integrals ?
 A: This is not true even for continuous nonnegative $f.$ For $N=1,2,\dots,$  we define $f$ on the disjoint intervals $[N-1/(N^2e^N),[N+1/(N^2e^N)]$ to be an isosceles triangular spike of height $e^N$ centered over this interval. Define $f$ to be $0$ everywhere else. Then $f$ is continuous everywhere, and
$$\int_{-\infty}^{\infty}f(x) \ dx = \sum_{N=2}^{\infty}1/N^2 < \infty.$$
But for each $N,$
$$\sum_{m \in\mathbb{Z}} \frac{1}{N} f \left( \frac{m}{N}\right) > \frac{1}{N}f \left( \frac{N^2}{N} \right) =  \frac{e^N}{N}\to \infty$$
as $N\to \infty.$
A: With the assumption that $f$ is continuous, this holds. WIthout the assumption of continuity it does not in general. For example $$ f(x) = \begin{cases} e^{-x^2} & x - z \in \Bbb Q\\0 & x - z \notin \Bbb Q\end{cases}$$ The limit is positive, but the integral is $0$.
Edit:
As zhw has shown, the expression does not necessarily hold even for continuous $f$. But there are more ways to rescue it than just functions that are monotonic on either side of some point. One trick is simply to reverse the limits: Exchange the limits on $m$ and $n$ and the result holds for continuous functions:
$$\lim_{M\to \infty} \lim_{n \to \infty} \sum_{m = -M}^M \frac 1 n f\left (\frac{m + z}{n}\right ) = \int_{-\infty}^\infty f(x)\ dx$$
As I noted in the comments, $f$ is Riemann integrable on finite intervals because it is continuous. The limit on $n$ is just a Riemann sum for the integral $\int_{-M}^M f(x)\ dx$, and therefore must converge to it. Taking the limit as $M \to \infty$ then converges to the full integral over all of $\Bbb R$.
