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  "body": "\n## Counterexample\n\nLet $n \\ge 2$ and let $\\Omega = \\set{x \\in \\R^n : 0 \u003c \\abs{x} \u003c 1}$ be the unit [[open ball|0YE5868HZB]] with its centre removed.\nIts boundary is the unit sphere together with the point $0$.\nLet $f = 0$ on the unit sphere and $f(0) = 1$; this $f$ is continuous on $\\partial\\Omega$, since $0$ is an isolated point of $\\partial\\Omega$.\nThe [[Perron–Wiener–Brelot solution|R12KQGVGN6]] of $f$ is $H_f = 0$, so $0$ is not a [[regular boundary point|B7948M365K]].\nCheck: the constant $0$ is in the lower class, so $H_f \\ge 0$.\nFor $\\varepsilon \u003e 0$, let $u_\\varepsilon(x) = \\varepsilon (\\abs{x}^{2-n} - 1)$ if $n \\ge 3$ and $u_\\varepsilon(x) = \\varepsilon \\log (1/\\abs{x})$ if $n = 2$.\nThis function is harmonic, hence [[superharmonic|X4V2S56YQQ]], on $\\Omega$.\nIt is $0$ on the unit sphere and tends to $+\\infty$ at $0$, so it is in the upper class and $H_f \\le u_\\varepsilon$.\nLetting $\\varepsilon \\to 0$ gives $H_f \\le 0$, so $H_f = 0$, which does not tend to $f(0) = 1$.\n\n## Remarks\n\nSo the [[Dirichlet problem|8D56GQ4AG8]] on the punctured ball has no solution for this continuous boundary function.\nA classical solution $u$ would be bounded and harmonic near $0$, so $0$ would be a removable singularity.\nThe extended function on the ball would vanish by the maximum principle, which contradicts $u(0) = 1$.\nThe point $0$ is too small to be seen by harmonic functions: it has capacity zero.\n\n## Explanations\n\n### L1 human\n\nTake a ball, keep the surface at temperature $0$, and try to hold the single centre point at temperature $1$.\nA single point cannot hold heat in equilibrium, so the temperature in the ball is $0$ everywhere.\nThis shows that the Dirichlet problem can fail for a very simple open set.\n",
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