Regularity of $\phi$ in order that $\int g_h \phi \,dx \to \phi(0)$ Define the sequence of functions $(g_h)_h$ where $$g_h(x):= h\, \chi_{[0,1/h]}(x)$$ and the sequence of measures $$(\mu_h(dx))_h:= g_h(x)\,dx.$$
We want to show that $\mu_h  \stackrel{*}{\rightharpoonup} \delta_0$. So take $\phi\in C^0_c(\mathbb R)$ and consider $\int_\mathbb R g_h(x) \phi(x) \,dx.$ We have to prove that it tends to $\phi(0)$ as $h\to \infty$.
Firstly, I've done something and I want to ask you if it is correct.
I see that $\int_\mathbb R g_h(x) \phi(x) \,dx$ is exactly the mean of the function $\phi$ over the interval $[0,1/h]$, so I know that for each $h$ there exists an $x_h\in[0,1/h]$ such that $\int_\mathbb R g_h(x) \phi(x) \, dx=\phi(x_h)$. Now clearly $x_h\to 0$, so, since $\phi$ is continuous, $\phi(x_h)\to \phi(0)$, proving the statement.
Secondly, I actually have used the continuity of $\phi$, but is it really necessary? I mean, is it true that $\int_\mathbb R g_h(x) \phi(x)\, dx\to \phi(0)$ even if $\phi$ as a lower regularity? For example $\phi \in L^1.$
 A: You are right that the sequence $g_h$ you have here has a distinct behaviour when it is paired with (almost) arbitrary test functions $\phi$, but you have to be careful in describing it. As it is usual in measure theory, functions are defined up to "almost-everywhere equivalence", so their pointwise behaviour at a point is ill-defined. This means that the statement 
\begin{equation}
\int_{\mathbb{R}}g_h\phi\, dx \to \phi(0)
\end{equation}
simply does not make sense in general. What does make sense is the following statement: 
\begin{equation}\tag{1}
\phi\ast g_h \to \phi,\qquad \text{in some sense}, 
\end{equation}
where the $\ast$ symbol is called "convolution" and is defined by 
$$\phi\ast g_h\,(x)=\int_{\mathbb{R}} g_h(x-y)\phi(y)\, dy.$$
Turns out that $(1)$ is true in quite a general way. The convergence holds in pointwise sense (at almost all points, or at all points if $\phi$ is continuous) and also in pretty much any functional analytic setting you wish. For example, if $\phi\in L^p$ then the convergence holds in $L^p$ sense (except for the endpoint case $p=\infty$ which is more delicate).
See here for more information: 
http://en.wikipedia.org/wiki/Approximation_to_the_identity
