Let $\delta_0$ be the measure defined as $\delta_0(A)=1$ if $0 \in A$ and $\delta_0(A)=0$. Let $g_h(x)=\frac{1}{h}1_{[-h,h]}$ and define $\nu_h$ as:
$\nu_h(A)=\int_{A}g_h(x)dx$.
Show that for every $f$ continuous we have
$lim_{h \to 0}\int_{\mathbb{R}}fd\nu_h=\int_{\mathbb{R}}fd\delta_0$.
My attempt:
So, I think it's a tricky application of Lebesgue Differentiation Theorem which states that(as in folland): If $k\in L^1$ then for almost every x:
$lim_{r \to 0} \frac{1}{m(E_r)}\int_{E_r}k(y)dy=k(x)$
for every family $E_r$ that shrink nicely to x.
So for us, $E_r=[-h,h]$, then
$\nu_{h}(A)=\int_{E_r}g_h(x)dx=\int_{E_r}\frac{1}{2h}1_{[-h,h]}dx=\frac{1}{2h}\mu(E_r\cap[-h,h])=\frac{2h}{2h}=1$,
and then I don't know how to get the function $k$ to apply LDT.
Thanks for any help!
For each $f\in L^1(\mathbb R)$, we have $\int fd\nu_h=\int fg_hdx=\frac{1}{h}\int_{-h}^{h}fdx.$
Applying the Lebesgue Density Theorem, we get
$\lim_{h\to 0}\int fdv_h=f(0).$
On the other hand, it follows from the definition of $\delta_0$ that
$\int fd\delta_0=\int\chi_{(-1/n,1/n)}\cdot fd\delta_0$
and since $\chi_{(-1/n,1/n)}\cdot f\le |f|,$
the Dominated Convergence Theorem gives
$\int fd\delta_0=\lim\int fd\delta_0=\lim\int\chi_{(-1/n,1/n)}\cdot fd\delta_0=\int\lim\chi_{(-1/n,1/n)}\cdot fd\delta_0=f(0)\cdot \delta_0(0)=f(0)$
and we conclude that $\lim_{h\to 0}\int fdv_h=\int fd\delta_0.$