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I used the same title as in Theorem 6.13 from Walter Rudin's Real and Complex Analysis, becouse I have the same question. However, in the link above I didn't find any proof for my question and I tried to prove it myself, but I can't. Any hints/suggestions? I am very thankful for the help.

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There seems to be a perfectly good proof in the comments under the question that you linked to. Since $h$ has absolute value 1, $\bar hg$ has the same absolute value as $g$: $|\bar h(x)g(x)|=|g(x)|$ for all $x$. In addition, for $\mu$-almost all $x$, we have that $\bar h(x)g(x)\geq0$ and therefore $\bar h(x)g(x)=|\bar h(x)g(x)|$. Combine those equations to get $\bar h(x)g(x)=|g(x)|$ for $\mu$-almost all $x$, as claimed.

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