Let $L_x^t$ local time of a standard Brownian motion on $(\Omega,\mathcal{F},\mathbb{P})$. Furthermore let $L_x^t$ the jointly continous version. Assume $m[0, \infty) \rightarrow [0,\infty)$ is increasing and right-continuous. Then
$t \mapsto A_t:=\int_{\mathbb{R}}L_x^t dm(x) $ is continuous $\mathbb{P}$-a.s.
In https://books.google.de/books?id=IlcOkvivjX0C&printsec=frontcover&dq=Dirichlet+Forms+and+Stochastic+Processes:+Proceedings+of+the+International&hl=de&sa=X&ved=0ahUKEwiP3uTI68HZAhVPJlAKHdKACcUQ6AEIJzAA#v=onepage&q=additive&f=false on site 120 (6) this statement is claimed. Is there are an easy proof?
My idea: Because $t \mapsto L_x^t $ is increasing for all $x$ a.s. I know $A_t$ is also increasing. Moreover there are only countable many jumps. If $m$ is atomic measure one has
$A_t= \sum_{i=1}^n p_i L_{x_i}^t$ which is clearly a.s. continuous.
The proof is quite obvious :). Using that $(x,t) \mapsto L_x^t$ is jointly continuous and that continuous functions on compact set(in $\mathbb{R}^n$ here $n=2$) are uniformly continuous.
Let $\delta>0$ and denote $Max:=\max_{y \in [0,s+\delta]}B_t$,$Min:=\min_{y \in [0,s+\delta]}B_t$. Then $(x,t) \mapsto L_x^t$ is uniformly continuous on
$[Min,Max] \times [0,s+\delta]$.
Furtheremore we have for all $t \in [0,s+\delta], x \in \mathbb{R}\setminus [Min,Max]$: $L_x^t=0$ This follows from $L_x^t = \lim_{a \rightarrow 0} \dfrac{1}{2 a} \int_0^t \textbf{1}_{[-a,a]}(B_s-a)ds$.
Therefore for all $\varepsilon>0$ exists $\delta'>0$ such that
$|L_x^t-L_x^r|<\varepsilon$ for all $(x,t) \in [Min,Max] \times [0,s+\delta]$ with$|(x,t)-(x,r)|< min(\delta,\delta')$
For $|t-r|<min(\delta,\delta')$ we obtain $|A_t-A_r| \leq \int_{[Min,Max]}|L_x^t-L_x^r| dm(x) \leq m([Min,Max]) \varepsilon $