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!


2 Answers 2


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}

Hint: consider Riemann sums

  • 1
    $\begingroup$ I think that the claim the OP wants to show is false. Actually, we should get $\int_0^1 f \, d\mu_n \to f(0)$. Note that $1/n$ or even $1/i$ is very different from $i/n$. In the last case, we would get $\int_0^1 f \, dx$. $\endgroup$
    – PhoemueX
    Commented Nov 12, 2015 at 21:46
  • $\begingroup$ Oh you're totally right. I knew something was off but I couldn't quite place it. Thanks for that. $\endgroup$ Commented Nov 12, 2015 at 21:56
  • $\begingroup$ Yeah I'm sorrry it's $i/n$. Now edited in the problem $\endgroup$
    – Ninanina
    Commented Nov 12, 2015 at 22:36
  • $\begingroup$ @mathwizkid It happens! No worries. I edited my answer to give you a good idea of how to proceed. $\endgroup$ Commented Nov 12, 2015 at 23:08

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