Finding a measure $\mu$ so that $\int_\mathbb{R} \phi d\mu_n \to \int_\mathbb{R} \phi d\mu$ I'm try to solve the following problem, but so far I'm sort of at a loss on how to approach it:

Let $f_n(x) = n \cdot 1_{(0,1/n)}$ and $\mu_n(E) = \int_E f_n(x) dx$. Find a measure $\mu$ so that $$\int_\mathbb{R} \phi d\mu_n \to \int_\mathbb{R} \phi d\mu$$as $n \to \infty$ for every continuous function in $\mathbb{R}$.

Any hints for getting started would be greatly appreciated!
 A: It is fairly easy to see that $\mu_n(E) = n \times \mu(E \cap (0,\frac 1n))$. Using approximations, it is not difficult to show that for any continuous $\phi$ we have $\int_{\Bbb R} \phi d\mu_n = n\int_{0}^\frac 1n \phi dx$ where $dx$ is the usual Lebesgue measure.
However, we have the mean value theorem for integrals which requires continuity of $\phi$ : this tells us that $n \int_0^\frac 1n \phi dx = \phi(\xi)$ for some $\xi \in (0,\frac 1n)$. Now, it is clear that the limit of the desired quantity is $\phi(0)$, since $\phi(\xi)$ can be brought close enough to $\phi(0)$ by increasing $n$, since $0 < \xi < \frac 1n$.
In other words, using $\phi$ to approximate indicator functions tells us that $\mu(E) = 1$ if $0 \in E$ else $\mu(E) = 0$. This is called the delta measure at $0$.
Note that $f_n$ converges to $0$ pointwise as a sequence of functions, so pointwise convergence of distribution functions differs from the "weak" convergence displayed here.
A: Let $g$ continuous.
Then $g$ is continuous at zero

Let $N \in \Bbb{N}$ and $h=1_E$ where $E$ is Lebesgue measurable, and denote $m$ the Lebesgue measure.
Then $\int_{\Bbb{R}}h d\mu_N=\mu_N(E)=N m(E \cap (0,\frac{1}{m}))=\int_{\Bbb{R}}N1_E(x)1_{(0,\frac{1}{N})}(x)dx=\int_{\Bbb{R}}h(x)f_N(x)dx$
By linearity of the integrals this is true for simple functions.
Now for every non-negative measurable functions $f$ we know that exists an increasing sequence of non-negative simple functions $h_n$ such that $h_n \to f$ pointwise.
By taking the negative and positive part of $g$ and using the above facts you have the identity $$\int_{\Bbb{R}}g(x)d\mu_N(x)=\int_{\Bbb{R}}g(x)f_N(x)dx$$

Let  $\epsilon>0$ then exists $\delta>0$ such that $|g(x)-g(0)|<\epsilon, \forall x\in (0,\delta)$
Exists $N \in \Bbb{N}$ such that $\frac{1}{n}<\delta, \forall n \geq N$
By the identity we proved before,we have:
$$|\int g d\mu_n-g(0)| \leq n\int_{(0,\frac{1}{n})}|g(x)-g(0)|dx < \epsilon$$
so  the limit is $g(0)$
This is true for every continuous functions thus for every continuous functions with compact support.
Since we can approximate indicator functions of sets with continuous  functions with compact support,then by approximation   the measure you seek is $\delta_0$
A: What you need to show is
$$\lim_{n\rightarrow \infty} \int_{\mathbb{R}}\phi\, d\mu_n = \phi(0).$$
Here the continuity of $\phi$ is essential.
Can you now find a $\mu$ for what you want?
