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Let $f$ be a Lebesgue integrable function. A point $x$ in the domain of $f$ is a Lebesgue point if $f(x)=\lim_{r\to 0}\frac{1}{2r}\int_{x-r}^{x+r} f(y) dy$. How can I prove that $\sin x$ and $\cos x$ have common Lebesgue point in (0;1)?

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Suppose $f\colon \mathbb R\to \mathbb R$ is locally integrable and continuous at $x$. Then $x$ is a Lebesgue point of $f$: Let $\epsilon > 0$, then there is a $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ for $|x-y| < \delta$. For $r < \delta$ we have \begin{align*} \left|f(x) - \frac 1{2r}\int_{x-r}^{x+r} f(y)\, dy\right| &= \left|\frac 1{2r}\int_{x-r}^{x+r} f(x)- f(y)\, dy\right|\\ &\le \frac 1{2r}\int_{x-r}^{x+r} \left|f(x)- f(y)\right|\, dy\\ &\le \epsilon. \end{align*} As $\sin,\cos\colon (0,1) \to \mathbb R$ are continuous each point is a common Lebesgue point.

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Since $sin(x)$ and $cos(x)$ are continuous functions every point in $(0;1)$ is a Lebesgue point. Indeed you can use the mean value property and the continuity of $f$ and get $ \exists \theta(x, r)\in [-1, 1]\qquad\frac{1}{2r}\int_{x-r}^{x+r}f(y)\,dy = f(x+r\theta(x, r))\rightarrow f(x) \quad\text{as}\quad r\rightarrow 0. $