For any $L>0$, let $\mu_L$ be the real Borel measure defined as follows:
$$
\mu_L=\frac{1}{L}\sum_{x\in\mathbb Z}\delta_{x/L},$$
$\delta_{x}$ being the atomic (Dirac) measure on $x$, and let $L^p(\mathbb R,\mu_L)$ the space of $p$-integrable complex-valued functions on the real line. By construction, for any $u$ in this space, we have
$$
\int_\mathbb{R}|u|^p\;\mathrm{d}\mu_L=\frac{1}{L}\sum_{x\in\mathbb{Z}}\left|u\left(\frac{x}{L}\right)\right|^p.$$
Spaces with different values of $L$ are isometric, as can be readily proven by dilation.
Now, the measure $\mu_L$ seems to "approach" the Lebesgue measure on the real line as $L$ becomes large: indeed, the integral above becomes a (improper) Riemann sum for $u$.
Does $\mu_L$ converge to the Lebesgue measure in some sense (e.g. weakly with respect to a space of suitably well-behaved functions)? I do have seen similar questions around for probability measures, but I do not know if such proofs can be extended for the Lebesgue measure on the full line.
Besides, if so, are $L^p(\mathbb{R},\mu_L)$ and $L^p(\mathbb{R})$ (the latter to be intended as the space of $p$-integrable functions with respect to the Lebesgue measure on $\mathbb{R}$) isometric?
Thanks in advance.
Best Answer
As far as I know, weak convergence is usually only defined for finite measures (of the same total mass). But for any compact set $K\subset\mathbb R$, it is indeed true that $\mu_L(\,\cdot\,\cap K)$ converges weakly to $\lambda(\,\cdot\,\cap K)$: If $f\colon K\to \mathbb R$ is continuous, $\int_K f d\mu_L$ is a corresponding Riemann sum.
Consequently, you have $\int_{\mathbb R} f d\mu_L \to \int_{\mathbb R} f d\lambda$ for all continuous functions $f\colon\mathbb R\to\mathbb R$ with compact support.