Let $(X,d)$ be a metric space and $f\in\text{Lip}_{\text{loc}}(X)$. We define the lower Lipschitz constant of u as the function $\text{lip }f:X\rightarrow[0,+\infty)$ given by:
$$\text{lip }f(x)=\liminf_{r\to 0^+}{\dfrac{\sup_{y\in B(x,r)}|f(y)-f(x)|}{r}}.$$
My question is how to prove that this function is Borel. This is what I have so far:
Fixed $\alpha\in\mathbb{R}$. If $\alpha<0$, it is obvious that $\text{\{lip $f\leq\alpha$\}}=\varnothing\in\mathcal{B}_X$. Suppose $\alpha\geq 0$. Fixed $x\in X$, we have:
Then:
$$\text{\{lip $f\leq\alpha$\}}=\bigcap_{n=1}^{\infty}{\bigcup_{0<r<1/n}{\bigcap_{y\in B(x,r)}{f^{-1}([f(y)-\alpha\cdot r,f(y)+\alpha\cdot r])}}}.$$
Since $f$ in continuous, $f^{-1}([f(y)-\alpha\cdot r,f(y)+\alpha\cdot r])$ is closed. The arbitrary intersection of closed sets is a closed set. So, $\bigcap_{y\in B(x,r)}{f^{-1}([f(y)-\alpha\cdot r,f(y)+\alpha\cdot r])}$ is closed.
I cannot assure now that at least $\bigcup_{0<r<1/n}{\bigcap_{y\in B(x,r)}{f^{-1}([f(y)-\alpha\cdot r,f(y)+\alpha\cdot r])}}\in\mathcal{B}_X$.