Let $\delta_a$ be the Dirac measure and $\mu_n = \frac{1}{n} \sum_{i=1}^n \delta_{i/n}.$ Show that $\lim_{n\to \infty} \int_0^1 f(x) \mu_n(dx) = \int_0^1 f(x) dx$ and if this is true, does it imply that $\mu_n(E) \to m(E)$ for every Borel measurable set?
The thing I am most confused about is $\mu_n(dx)$. I really don't get what this amount to and what kind of interpretation that this sequence of measure, $\mu_n(dx)$ has.
Thank you so much!
On
Integrating against $\mu_n$ is the same as using the right hand rectangle rule with $n$ points, as it would be used to estimate $\int_0^1 f(x) dx$. For Riemann integrable functions, this will always converge to $\int_0^1 f(x) dx$. For general Lebesgue integrable functions this is not true; for instance if $f=1_{\mathbb{Q}}$ then the right hand rectangle rule always gives us $1$ but the Lebesgue integral is zero. So some hypotheses are required to get your statement.
Measure theory has slightly weird and very inconsistent notation at times. Personally I take the approach that measures are denoted $d\mu(x)$ when treated as integration variables. It tells you: 1) you're integrating, 2) which measure you're integrating with respect to, and 3) which variable you're integrating with respect to. The notation $\mu(dx)$ is clunky at best. I'll get you started so you understand the problem. You can finish where I leave off.
\begin{align} \int_0^1 f(x)\,d\mu_n(x) &= \int_0^1 f(x)\left(\frac{1}{n}\sum_{i=1}^n d\delta_{\frac{i}{n}}(x)\right) \\ &= \frac{1}{n}\sum_{i=1}^n \int_0^1 f(x)\,d\delta_{\frac{i}{n}}(x) \\ &= \frac{1}{n} \sum_{i=1}^n f\left(\frac{i}{n}\right). \end{align}