Uniform convergence of proportion of positive real numbers at which uniformly convergent sequence of functions is positive For a continuous function $g: \mathbb R \rightarrow \mathbb R$, we define the function $p_g: \mathbb R^+ \rightarrow \mathbb R^+$ by
$$p_g(T) := \frac 1T\lambda(g^{-1}((0, \infty)) \cap [0, T])$$
where $\lambda$ denotes the Lebesgue measure.
Let $f_n: \mathbb R \rightarrow \mathbb R$ be a sequence of continuous functions which uniformly converges to a continuous function $f: \mathbb R \rightarrow \mathbb R$. What I am interested in finding out is whether the sequence of functions $\{p_{f_n}\}_{n=1}^\infty$ converges uniformly to $p_f$ on say, some $[A, \infty)$ for some fixed positive real number $A$. If necessary, we may assume the $f_n$'s and $f$ to be smooth.
Writing
$$\lambda(g^{-1}((0, \infty)) \cap [0, T]) = \int_0^T \mathbbm{1}(g(t)>0) \, \text dt,$$
the Dominated Convergence Theorem shows that $p_{f_n}(T)$ converges to $p_f(T)$ pointwise for each $T>0$. However, I am having some trouble getting the uniform convergence I want. It may be useful to note that $p_g(T)$ is just the probability of $x \in [0, T]$ of having $p_g(x)>0$.
I feel like this should be fairly straightforward and should hold in greater generality (say, without any assumptions besides the uniform convergence $f_n \rightarrow f$ and, perhaps, Lebesgue measurability of the functions $f_n$ and $f$). I fear I may be missing some really simple idea; I haven't found any result of this form anywhere either, even though it looks fairly standard. Hence, I would be really grateful for any suggestions or references, even if one needs to assume that $f_n$'s and $f$ are smooth.
P.S.: In addition, what are the minimal conditions on the $f_n$ and $f$ for this to hold?
 A: I have some results here: First of all, I am going to write $g^{-1}((0,\infty))$ as $\{g>0\}$.
Second of all, I don't think mere convergence of $f_n$ to $f$ is enough to use Dominated Convergence to show that $p_{f_n}\to p_f$. As you correctly point out you can rewrite $$p_g(T)=\frac{1}{T}\int \unicode{x1D7D9}_{[0,T]\cap\{g>0\}}d\lambda$$ and the sequence is obiviously dominated by $$\unicode{x1D7D9}_{[0,T]}$$ which is obviously integrable, but in order to use Dominated Convergence on this form you would have to show that $$\unicode{x1D7D9}_{[0,T]\cap\{f_n>0\}}\to\unicode{x1D7D9}_{[0,T]\cap\{f>0\}}$$ pointwise first.
Let $x\in[0,T]$. Then $$|\unicode{x1D7D9}_{[0,T]\cap\{f_n>0\}}(x)-\unicode{x1D7D9}_{[0,T]\cap\{f>0\}}(x)|$$ is either $0$ or $1$. If $f(x)>0$ and $f_n$ converges pointwise to $f$ then there exists an $n$ s.t. $f_n(x)>0$ and the absolute difference is therefore $0$. Same deal if $f(x)<0$. However, if $f(x)=0$ then the absolute difference might always be $1$ if the convergence is from above, right? $f_n(x)$ could always be positive, but converge to $0$.
So, all you need is, if $f$ is measurable and $f_n\to f$ pointwise such that the zeros of $f$ are approached from below then $p_{f_n}\to p_f$ pointwise.
Now, notice that for any measurable function $g$, $p_g$ is continuous. If you look at the function $$L(t)=\lambda(A\cap[0,t])$$ for some set of real numbers $A$, you get for any $x>y>0$ $$|L(x)-L(y)|=|\lambda(A\cap[y,x])|\leq x-y$$ which means it is even Lipschitz-continuous. Using $A=\{g>0\}$ and multiplying by the multiplicative inverse function, gets you $p_g$ which is also continous since those functions are closed under multiplication.
Meaning, if $f$ is measurable and $f_n$ is a monotonically increasing sequence which converges pointwise to $f$, we get by the result from before that $p_{f_n}\to p_f$ pointwise. Furthermore, the sequence $p_{f_n}$ is also monotonically increasing and $p_{f_n}$ and $p_f$ are continuous and therefore by Dini's Theorem $p_{f_n}\to p_f$ compactly.
