Assume that $(\nu_{j})$ is a sequence of positive Radon measures supported in a compact set $K\subseteq\mathbb{R}^{n}$ and \begin{align*} \sup_{x\in{\mathbb{R}}^{n},r>0}\dfrac{1}{r^{n-\lambda}}\int_{B(x,r)}|M_{\alpha}\nu_{j}(y)|^{p}dy\rightarrow 0, \end{align*} where $0<\lambda,\alpha<n$, and \begin{align*} M_{\alpha}\nu_{j}(y)=\sup_{s>0}\dfrac{1}{s^{n-\alpha}}\nu_{j}(B(y,s)). \end{align*} The author claims that there is a subsequence of $(\nu_{j})$ that $w^{\ast}$ converges to a measure $\nu$ which supported in $K$.
I fail to see why it must be the case.
Apparently the author resorts to the Banach-Alaoglu Theorem of the sequential $w^{\ast}$ compact property regarding to the space of all positive Radon measures which supported in $K$.
Therefore we are to bound $(\nu_{j}(K))$. But how does one bound the quantity $\nu_{j}(K)$ in terms of the integral with respect to the fractional maximal function?
Assume that $K\subseteq B(0,M)$ for some $M>0$. Then for $y\in B(0,1)$ we have $B(y,M+1)\supseteq K$ and hence \begin{align*} \sup_{x\in{\mathbb{R}}^{n},r>0}\dfrac{1}{r^{n-\lambda}}\int_{B(x,r)}|M_{\alpha}\nu_{j}(y)|^{p}dy&\geq\int_{B(0,1)}|M_{\alpha}\nu_{j}(y)|^{p}dy\\ &\geq\int_{B(0,1)}\left(\dfrac{1}{(M+1)^{n-\alpha}}\nu_{j}(B(y,M+1))\right)^{p}dy\\ &\geq\dfrac{1}{(M+1)^{p(n-\alpha)}}\int_{B(0,1)}\nu_{j}(B(y,M_1))^{p}dy\\ &\geq\dfrac{1}{(M+1)^{p(n-\alpha)}}\int_{B(0,1)}\nu_{j}(K)^{p}dy\\ &=\dfrac{v_{n}}{(M+1)^{p(n-\alpha)}}\nu_{j}(K)^{p}, \end{align*} so $\lim_{j\rightarrow\infty}\nu_{j}(K)=0$ and hence the sequence $(\nu_{j}(K))$ is bounded.