Show that $ \mu_n \to \mu $, in the weak* topology, where $ \mu \ll \nu $ and $ \dfrac{d\mu }{d\nu}=\dfrac{f}{\|f\|_1} $

Question

Let $ (X,\mathcal{B}_X,\nu) $ be a measure theoretical space, where $ X $ is a compact metric space, $ \mathcal{B}_X $ its Borel and $ \nu $ a probability measure. Let $ \{ f_n \}_n\in\mathbb{N} $ be a sequence of functions defined on $ X $ such that each $ f_n\in L^1(\nu) $ and $ f_n\to f $ in 1-mean, i.e., $ \lim\limits_{n\to \infty}\|f_n- f\|_1 =0$. Moreover, assume that $ \|f\|_1>0 $. For each $ n\in \mathbb{N} $ define the measure $ \mu_n $ such that $ \mu_n \ll \nu $ and $ \dfrac{d\mu_n}{d\nu}=\dfrac{f_n}{\|f_n\|_1} $. Show that $ \mu_n \to \mu $, in the weak* topology, where $ \mu \ll \nu $ and $ \dfrac{d\mu }{d\nu}=\dfrac{f}{\|f\|_1} $

Anyone?

Answer 1

For every continuous test function $\phi$, by definition we have $$\int\phi(x)d\mu_n(x)=\int\phi(x)\frac{f_n(x)}{\|f_n\|_1}d\nu(x)$$ and $$\int\phi(x)d\mu(x)=\int\phi(x)\frac{f(x)}{\|f\|_1}d\nu(x).$$ Therefore $$|\langle\mu_n,\phi\rangle-\langle\mu,\phi\rangle| =\left|\int\phi(x)\left(\frac{f_n(x)}{\|f_n\|_1}-\frac{f(x)}{\|f\|_1}\right)d\nu(x)\right|\leq\|\phi\|_\infty \left\|\frac{f_n(x)}{\|f_n\|_1}-\frac{f(x)}{\|f\|_1}\right\|_1$$ and the last norm goes to zero because $\|f_n\|_1\to\|f\|_1$ and \begin{align} \left\|\frac{f_n(x)}{\|f_n\|_1}-\frac{f(x)}{\|f\|_1}\right\|_1 &\leq \left\|\frac{f_n(x)}{\|f_n\|_1}-\frac{f(x)}{\|f_n\|_1}\right\|_1+\left\|\frac{f(x)}{\|f_n\|_1}-\frac{f(x)}{\|f\|_1}\right\|_1\\ &=\frac{1}{\|f_n\|_1}\|f_n-f\|_1+\|f\|_1\left\|\frac{1}{\|f_n\|_1}-\frac{1}{\|f_n\|_1}\right\|_1\to 0. \end{align}