I'm trying to prove the following criteria for weak compactness of Radon measures:
Let $\{\mu_k\}_{k \in \mathbb{N}}$ be a sequence of radon measures on $\mathbb{R}^n$ satisfying $$\sup_{k \in \mathbb{N}}\mu_k(K)<\infty$$ for each compact set $K\subset \mathbb{R}^n$. Then there exists a subsequence $\{\mu_{k_j}\}_{j \in \mathbb{N}}$ and a radon measure $\mu$ such that $\mu_{k_j}$ converge weakly to $\mu$.
In the proof I found in Evans & Gariepy, the authors claim there is no loss of generality in supposing that $\displaystyle{\sup_{k\in \mathbb{N}}\mu_k(\mathbb{R}^n)} < +\infty$, since otherwise we could work with the measures $\mu_{k}^{\ell} = \mu_k \vert_{B(0, \ell)}$ and conclude the proof by using a diagonal argument. This is a little hazy for me. Sure, we can conclude that for each $k \in \mathbb{N}$ there is a convergence subsequence $\left\{\mu_{k}^{\ell_j}\right\}_{j \in \mathbb{N}} $ of $\left\{\mu_{k}^{\ell} \right\}_{\ell \in \mathbb{N}}$, but how does one get rid of the "$\ell$-dependency" in order to extract a convergent subsequence $\left\{ \mu_{k_j}\right\}_{j \in \mathbb{N}}$ of $\{\mu_k\}_{k \in \mathbb{N}}$?
The diagonal argument works like this: You take a subsequence of $\langle \mu_m^1\rangle$ of $\langle\mu_k\rangle$ whose restriction to $B_0(1)$ converges weakly. Then you take a further subsequence $\langle\mu^2_n\rangle$ of $\langle\mu_m^1\rangle$ (and, thus, also $\langle\mu_k\rangle$) whose restriction to $B_0(2)$ converges weakly. Continue this way to get a sequence $\langle \langle\mu_p^l\rangle\rangle_{l=1}^\infty$ of subsequences. Then the sequence $\mu_1^1,\mu_2^2,\mu_3^3,\ldots$ is eventually a subsequence of each of these subsequences, and converges, therefore, weakly when restricted to any ball around $0$. Since every compactly supported function vanishes outside some such ball, the sequence must converge weakly.