If $W=\{W_t:t\geq0\}$ is a standard Brownian motion with quadratic variation $\langle W\rangle_t=t$ then we can measure the amount of time spent in an open set $A$ by $\int\mathbf{1}_A(W_t)\,\mathrm{d}t$. In particular, one can readily prove that, if $A$ is an open interval about $0$,
$$L^0_t(W)=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\int\mathbf{1}_{\varepsilon A}(W_t)\,\mathrm{d}t\tag{$\ast$}$$
where $L^0$ is the local time, a measure of time spent at level $0$. (Depending on definition of local time there may be a $1/2$ factor.) The generalization to continuous semimartingales $X$ correspondingly has
$$L^0_t(X)=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\int\mathbf{1}_{\varepsilon A}(X_t)\,\mathrm{d}\langle X\rangle_t.$$
Clearly in the case $X=W$ this gives me Brownian local time. What I'm having trouble understanding is why $\mathrm{d}\langle X\rangle_t$ is the preferred (signed) measure to help me control how much time a general semimartingale $X$ is spending in $A$. Is there a physical motivation for using quadratic variation as a measure here? Because surely the naive approach would be to consider $\int\mathbf{1}_A(X_t)\,\mathrm{d}t$, and in the limit for local time (ignoring generally problems of existence).
Some books take $(\ast)$ as the definition of local time. It's more classical to define it as the $L^0$ in
$$\mathrm{d}|X_t|=\operatorname{sgn}(X_t)\,\mathrm{d}X_t+\mathrm{d}L_t^0(X).$$
They are equivalent up to a $1/2$ constant.