# Making a non-convergent sequence of measures convergent by additive perturbation

Denote by $M$ the topological dual of the Banach space $C[0,1]$. Then $M$ can be identified with the space of signed measures on $[0,1]$. Denote by $M^+ \subseteq M$ the positive cone of $M$ consisting of positive measures. Each $\mu \in M$ has the Jordan decomposition $\mu = \mu^+ - \mu^-$ with $\mu^\pm \in M^+$ and set $|\mu| := \mu^+ + \mu^-$. Equip $M$ with the weak* topology such that a sequence $\mu_n \in M$ converges to $\mu$ iff $\int_0^1 f d\mu_n \to \int_o^1 f d\mu$ for all $f \in C[0,1]$.

Assume that $\mu_n \to 0$. Then $|\mu_n|$ does not need to converge ($|.| : M \to M^+$ is not continuous) but $|\mu_n|$ has a convergent subsequence (from $\mu_n \to 0$ it follows that the total variation norms $|| \, |\mu_n| \, || = || \mu_n ||$ are uniformly bounded and we can apply the Helly selection theorem on $|\mu_n|$).

Question. Is there a sequence $\sigma_n \in M^+$ (of positive measures) such that $|\mu_n| + \sigma_n$ converges weakly*?

Probably hopeless idea: remove the convergent subsequence of $|\mu_n|$, providing a (finite or infinite) sequence of remaining elements of $|\mu_n|$ which in turn has a convergent subsequence and proceed iteratively. If this iteration terminates after finitely many steps, we can cook up such a sequence $\sigma_n$ explicitely. But the problem is that this iteration does not need to stop in finitely many steps. So I think that this idea is hopeless and gives probably a hint that such a sequence $\sigma_n$ of positive measures need not exist.

This is not possible. Consider the following sequence. Let $(r_n)$ denote an enumeration of the rational numbers in $[1/4,3/4]$. Define $L^1$-functions $$\mu_n(x):= 2^{n-1}\left(\chi_{[r_n-2^{-n},r_n]}(x)-\chi_{[r_n,r_n+2^{-n}]}(x)\right)\in L^1(0,1), \ n\ge2.$$ Then $\mu_n\to 0$. Moreover, for each $x\in[1/4,3/4]$ there is a subsequence of $(|\mu_n|)$ converging to $\delta_x$.
Suppose the claim is true. Then $|\mu_n|+\sigma_n\to \nu$. Since $|\mu_n|+\sigma_n\ge |\mu_n|$ it follows that $\nu\ge \delta_x$ for all $x\in[1/4,3/4]$. Hence $\nu( (0,1))=+\infty$, which is impossible.
• Very nice counterexample. Thank you. (The factor $2^{n+1}$ should be changed to $2^{n-1}$ for $|\mu_{n_k}| dx \to \delta_x$). – yadaddy Aug 18 '16 at 5:56