I'm pretty much stuck on part (ii). Any guidance would be appreciated!

The Dirac measure $\delta_{x_0}$ for $x_0$∈$\mathbf{R}$ is defined over the measurable space $\left(\mathbf{R}, \mathbf{B}\right)$ by:

$\begin{equation} \delta_{x_0}(B)=\begin{cases} 1, & \text{if $x\in B$}.\\ 0, & \text{if $x\in B^C$}. \end{cases} \end{equation}$

(i) Show that $\delta_{x_0}$ is a probability measure for all $x\in \mathbf{R}$.

(ii) Show that, for every continuous function $f\colon \mathbf{R}\to [0,∞)$,

$\lim_{x \to \infty}\int_\mathbf{R} f d\delta_{1/n}= \int_\mathbf{R} f d\delta_0$

  • 1
    $\begingroup$ In (ii) it appears that you have specialized the setting to the case $\mathbf{R} = \mathbb{R},$ the set of Real numbers. Here are some hints. Evidently you will need to exploit the assumed continuity of $f$ and, almost as evidently, you really only need continuity in a neighborhood of $0$. Can you find a simple formula for $\int_\mathbf{R} f d\delta_{x}$ in terms of $f$ and $x$? $\endgroup$
    – whuber
    Nov 29, 2013 at 18:38

1 Answer 1


First, express the Dirac measure over $(\mathbb{R},\mathscr{R})$ concentrated at $x_0\in\mathbb{R}$ by means of an indicator: $$ \delta_{x_0}(A)=I_A(x_0) \, , $$ for $A\in\mathscr{R}$. Second, to prove that $\delta_{x_0}$ is indeed a (probability) measure you have to check that: i) $\delta_{x_0}(A) \in [0,1]$, for every $A\in\mathscr{R}$ (easy, it is an indicator); ii) $\delta_{x_0}(\emptyset)$ (easy, the empty set has no elements); iii) If $A_1,A_2,\dots$ is a disjoint sequence of sets in $\mathscr{R}$, then $$ \delta_{x_0}\left( \cup_{i=1}^\infty A_i \right) = \sum_{i=1}^\infty \delta_{x_0}(A_i) \, $$ (again easy, since the $A_i$'s are disjoint, $x_0$ can be in no more than one of them). For the other item, note that $$ \int_{\mathbb{R}} f(x)\,d\delta_{1/n}(x) = f(1/n). $$ If $f$ is continuous, what can you say about $\lim_{n\to\infty} f(1/n)$?


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