Lebesgue function of interpolation: is $\limsup L_n(x)/\log n \ge 2/\pi$ almost everywhere? (Erdős #1132)

For $x_1,\ldots,x_n\in [-1,1]$ let $$l_k(x)=\frac{\prod_{i\neq k}(x-x_i)}{\prod_{i\neq k}(x_k-x_i)}$$ be the Lagrange fundamental polynomials (so $l_k(x_k)=1$ and $l_k(x_i)=0$ for $i\neq k$). Let $x_1,x_2,\ldots\in [-1,1]$ be an infinite sequence of nodes and define the Lebesgue function $$L_n(x) = \sum_{1\leq k\leq n}\lvert l_k(x)\rvert,$$ where each $l_k$ is taken with respect to the first $n$ nodes $x_1,\ldots,x_n$. Must there exist $x\in (-1,1)$ such that $$L_n(x) >\frac{2}{\pi}\log n-O(1)$$ for infinitely many $n$? Is it true that $$\limsup_{n\to \infty}\frac{L_n(x)}{\log n}\geq \frac{2}{\pi}$$ for almost all $x\in (-1,1)$?

Acceptance. FULLY RESOLVES: a complete proof (machine-checkable Lean/Coq preferred, else a full written proof) of both parts — (a) for every infinite node sequence there exists $x\in(-1,1)$ with $L_n(x) > (2/\pi)\log n - O(1)$ for infinitely many $n$, and (b) $\limsup_n L_n(x)/\log n \geq 2/\pi$ for almost every $x\in(-1,1)$; OR a disproof — an explicitly constructed node sequence together with a proof that the relevant conclusion fails (e.g. a positive-measure set of $x$ with $\limsup L_n(x)/\log n < 2/\pi$). Because of the ambiguity noted in the background, a resolution of part (a) must state explicitly whether its $O(1)$ is uniform or $x$-dependent, and which reading it settles. ADVANCES: (a) upgrading the dense-set result stated in the background to a set of positive (or full) measure, even with a weaker error term $\omega(n)$; (b) proving part (a) in the $x$-dependent-constant reading; (c) establishing the conjectured bound for natural structured families of node arrays (e.g. all arrays with asymptotic distribution different from the arcsine measure), with proof. Deliver the proof file(s).