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Assume that $f$ is harmonic in the unit disk $|z|<1$, with boundary function of bounded variation, such that $$\lim_{r\to 1}f(re^{it})= 0$$ for $t\in[0,\pi]\setminus \mathbf{Q}$, where $\mathbf{Q}$ are rational numbers. Can we then state the following $$\lim_{r\to 1}f(re^{it})= 0,\ \ \ t\in[0,\pi]. $$

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What do you exactly mean by "boundary function"? –  Alexandre Eremenko Mar 9 '13 at 3:18
    
$f=P[g]$, where $g(t):[0,2\pi]\to \gamma\subset \mathbf{C}$ is of bounded variation. –  djoke Mar 9 '13 at 10:17
    
You can use the fact that Fourier series of a function of bounded variation converges everywhere to arithmetic mean of left and right limits. –  Mateusz Wasilewski Mar 9 '13 at 12:41

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