Convergence in total variation norm

Question

Let $\mu_n = \sin^2(n\pi x)dx$ be a sequence of measures on the compact interval $[0,1]$. Moreover, let $\mathcal{M}([0,1])$ be the Banach space of of bounded Borel measures on $[0,1]$ equipped with the total variation norm, $$\| \nu(E) \| = \sup \sum_{j=1}^{\infty} \left| \nu(E_j) \right|.$$ I want to show that $\mu_n$ is not a norm-convergent sequence. Since we are in a Banach space, Cauchy is equivalent to convergence. Hence I want to arrive at a contradiction by suppose that $\mu_n$ is a Cauchy sequence. Therefore, let $\epsilon >0$ and suppose that for $m,n \geq N$ we have $$\| \mu_n - \mu_m \| < \epsilon.$$ I have only begun looking at the total variation norm, and am unsure how to use it correctly, so I haven't been able to make progress beyond this point.

Answer 1

Hint: If $\mu = f \, \mathrm{d}x$, you have $\| \mu \|_{\mathcal M([0,1])} = \| f \|_{L^1(0,1)}$ (try to prove it!).