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Let $f$ be a complex-valued, measurable function defined on $\mathbb R$ with $f(x)=f(x+2\pi)$

I want to show $$\int_{[0,2\pi]}f d\lambda=\int_{[\alpha,\alpha+2\pi]}f d\lambda $$

$(\alpha \in \mathbb R)$

My original attempt was to write $\int_{[\alpha,\alpha+2\pi]}f d\lambda -\int_{[0,2\pi]}f d\lambda=\int_{\alpha}^{\alpha+2\pi}f(x)dx-\int_{0}^{2\pi}f(x)dx=F(\alpha+2\pi)-F(2\pi)-(F(\alpha)-F(0))=\int_{2\pi}^{\alpha+2\pi}f(x)dx-\int_{0}^{\alpha}f(x)dx=0$

But I think I can't simply write is a Riemann Integral.

How can I show $$\int_{[0,2\pi]}f d\lambda=\int_{[\alpha,\alpha+2\pi]}f d\lambda $$ instead?

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Edit: This proof assumes that $\alpha<2\pi$. But since $f$ is $2\pi$ periodic the function $f$ on $[\alpha,2\pi+\alpha]$ behaves the same as $f$ on $[\alpha-2\pi,\alpha]$ and we can keep removing $2\pi$ until $\alpha-2m\pi<2\pi$, then denote $\beta=\alpha-2m\pi$ and argue as below with $\beta$ instead of $\alpha$:

We can write $[\alpha,\alpha+2\pi] = [\alpha,2\pi]\cup [2\pi,2\pi+\alpha]$. Since the union is disjoint (up to one element) we have

$$\int_\alpha^{\alpha+2\pi} f(x)dx = \int_\alpha^{2\pi} f(x)dx+\int_{2\pi}^{2\pi+\alpha} f(x)dx$$

By assumption $f(x)=f(x+2\pi)$ and so by changing variables the second integral becomes

$$\int_{2\pi}^{2\pi+\alpha} f(x)dx = \int_0^\alpha f(x+2\pi)dx = \int_0^\alpha f(x)dx$$

Insert to the first equality we have

$$\int_\alpha^{\alpha+2\pi} f(x)dx = \int_\alpha^{2\pi} f(x)dx+\int_{0}^{\alpha} f(x)dx = \int_0^{2\pi}f(x)dx$$

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    $\begingroup$ What if $\alpha>2\pi$? How do you make sense of the interval $[\alpha,2\pi]$? $\endgroup$ – user587192 Jan 12 at 16:37
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    $\begingroup$ This would work if, instead of your "$2\pi$", you used the (unique) integer multiple of $2\pi$ between $\alpha$ and $\alpha+2\pi$. $\endgroup$ – paul garrett Jan 12 at 16:40
  • $\begingroup$ @user587192 If $\alpha>2\pi$ then move the entire interval by $2\pi$ as many times as needed. $\endgroup$ – Yanko Jan 12 at 16:54

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