• $a,b\in\mathbb R$ with $a<b$
  • $E$ be a $\mathbb R$-Banach space
  • $g:[a,b]\to E$ be of bounded variation and right-continuous

For any such $g$ let $\mu_g$ denote the unique (vector) measure on $\mathcal B((a,b])$ with $$\mu_g((s,t])=g(t)-g(s)\;\;\;\text{for all }a\le s\le t\le b\;.\tag1$$

Now, assume $g_n:[a,b]\to E$ is of bounded variation and right-continuous for $n\in\mathbb N$ such that $$\left\|g(t)-g_n(t)\right\|_E\xrightarrow{n\to\infty}0\;\;\;\text{for all }t\in[a,b]\tag2$$

I want to conclude $$\left\|\mu_g(A)-\mu_{g_n}(A)\right\|_E\xrightarrow{n\to\infty}0\;\;\;\text{for all }A\in\mathcal B((a,b])\;.\tag3$$

Clearly, $$\left\|\mu_g((s,t])-\mu_{g_n}((s,t])\right\|=\left\|(g(t)-g(s))-(g_n(t)-g_n(s))\right\|_E\xrightarrow{n\to\infty}0\tag4$$ for all $a\le s\le t\le b$ and hence $$\nu_g((s,t]):=\lim_{n\to\infty}\mu_{g_n}((s,t])\;\;\;\text{for }a\le s\le t\le b$$ is well-defined and coincides with $\mu_g$ on the semiring $$\mathcal H:=\left\{(s,t]:a\le s\le t\le b\right\}$$ which generates $\mathcal B((a,b])$.

A quick google search yielded the Vitali–Hahn–Saks theorem which seems to be applicable. However, above I've only shown convergence of $(\mu_{g_n}(A))_{n\in\mathbb N}$ for $A\in\mathcal H$, while the assumptions of the theorem include the convergence for all $A\in\mathcal B((a,b])$.

So, how can we show that convergence? And is there an elementary proof available which avoids the use of the mentioned theorem?


Convergence doesn't hold even for $E=\mathbb R.$ Let $[a,b]=[0,1],$ let $g=0$ and let $g_n(x)=1_{1-1/n\leq x< 1}(x)$ (the indicator function of $[1-1/n,1)$). All these functions are right-continuous and of bounded variation, and $g_n\to g$ pointwise. Let $A=\{1-1/2n\;\vert\;n\in\mathbb N\}.$ Then $\mu_{g_n}(A)$ is $1$ if $n$ is even and $0$ otherwise, so it doesn't converge.

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