on the boundary of analytic functions

The only way the function can be zero on an arc of the circle is for the function to be identically zero. To see this, consider ${\displaystyle g(z) = \prod_{j = 0}^{n-1} f(e^{2\pi i j \over n}z)}$ for a large enough $n$ so that $g(z)$ is identically zero on the boundary of the disk. Then by the Cauchy integral formula, if $r < 1$ then for any $w$ in the interior one has $$g(w) = {1 \over 2\pi i}\int_{|z| = r}{g(z) \over z - w}\,dz$$ Since $g(z)$ is continuous on the compact set $\bar{D}$, $g(z)$ is bounded, so one can apply the dominated convergence theorem and let $r$ go to $1$. One obtains $$g(w) = {1 \over 2\pi i}\int_{|z| = 1}{g(z) \over z - w}\,dz$$ $$ = 0$$ Thus $g(z)$ is identically zero. Since the zeroes of a nonzero analytic function are isolated, we conclude that $f(z)$ is identically zero as well.

Let $E$ be a closed subset of the boundary of the disk. Then there is a nonzero function $f$ continuous on the closed disk and analytic on the open disk such that $f\vert_E=0$ if and only if the Lebesgue measure of $E$ is zero.

The fact that the Lebesgue measure has to be zero (which answers your question) follows from a more general theorem that if $f$ is in any of the Hardy spaces of the disk, then the log of the absolute value of the boundary function of $f$ is in $L^1$ of the circle with Lebesgue measure. This is in Theorem 17.17 of Rudin's Real and complex analysis, 3rd Edition.

The fact that every closed subset of the boundary with Lebesgue measure zero can consist of zeros of a nonzero function that is continuous on the closed disk and analytic on the interior is a theorem of Fatou. This appears in the section "Theorems of Fatou and Rudin" in Chapter 6 of Hoffman's Banach spaces of analytic functions.

See also Myke's related question.