Measures converging weakly to Dirac delta function I have the following question: 
Let $\phi \in C_c^{\infty}([-1,1])$ be a smooth, compactly supported function with
integral $2$. Consider the sequence of measures on $[−1, 1]:$
$$\mu_n = n \phi(nx) dx$$ 
where $dx$ is Lebesgue measure. These measures converge in the weak-∗ topology.
What is the limit measure?
Truth is, I'm not entirely sure what I have to show. I know these functions converge to the Dirac delta function. I also know that computing something like 
$$\lim_{x\to 0^+} \int_{\mathbb R} n \phi(nx) dx$$
is maybe the right thing to do to show this...
I just can't put these together however. Could someone provide the first step or two just to start off, and I can take it from there? Thanks very much in advance. 
 A: What space you are looking at? $C(-1,1)$ or $\mathcal S(-1,1)$ or $C^\infty_C(\mathbb R)$ or what? Well anyway, $\mu_n\to\mu$ in the weak* sense iff $\mu_n(f)\to\mu(f)$ for all $f$ in whatever space you are living in.
At any rate your measures are $\mu_n=n\phi(nx)\, dx$. Note $\phi(x)$ is zero for $x$ outside of $[-1,1]$:
$$\mu_n(f)=\int_{\mathbb R}f(x)\, n\phi(nx)\, dx=n \int_{-1/n}^{1/n} f(x)\, \phi(nx)\, dx$$
Now $f$ should at the least be continuous. For that reason
$$\sup_{x\in[-1/n,1/n]}|f(x)-f(0)|\to0$$
as $n\to\infty$ (you actually only need continuity at $0$). So our integral can be written as:
$$\mu_n(f)=n\int_{-1/n}^{1/n} f(0) \phi(nx)\,dx+n\int_{-1/n}^{1/n}(f(x)-f(0))\phi(nx)\, dx$$
Now the first term evaluates to:
$$f(0) \int_{-1/n}^{1/n}n\phi(nx)\, dx = f(0)\int_{-1}^1 \phi(x) \, dx = 2f(0)$$
Whereas the second term can be bounded by:
$$\left|n\int_{-1/n}^{1/n}(f(x)-f(0))\phi(nx)\, dx\right|≤\sup_{x\in[-1/n,1/n]}|f(x)-f(0)|\cdot \|\phi\|\int_{-1/n}^{1/n}n dx$$
which converges to zero. So
$$\mu_n(f)\to 2f(0)$$
For all $f$. So $\mu_n\to 2\delta_0$.
