An exercise from Stein's Real Analysis about Lebesgue integral

Question

I would be glad if anyone check my solution for the following question.

Suppose $f$ is integrable on $(-\pi,\pi]$ and extended to $\Bbb R$ by making it periodic of period $2\pi$. Show that $\int_{-\pi}^\pi f(x)dx=\int_I f(x)dx$ where $I$ is any interval in $\Bbb R$ of length $2\pi$.

Attempt: Note that $I=(-\pi,\pi]+a=(-\pi+a,\pi+a]$ for some $a\in \Bbb R$. So, it is enough to show that for any $a\in \Bbb R$, $$\int_{-\pi}^\pi f(x)=\int_{-\pi+a}^{\pi+a}f(x)dx.$$ Here we go: \begin{align} \int_{-\pi+a}^{\pi+a}f(x)dx &=\int_{-\pi+a}^{\pi}f(x)dx+\int_\pi^{\pi+a}f(x)dx \\&=\int_{-\pi+a}^{\pi}f(x)dx+\int_{-\pi}^{-\pi+a}f(x+2\pi)dx \\&=\int_{-\pi+a}^{\pi}f(x)dx+\int_{-\pi}^{-\pi+a}f(x)dx, \quad \text{since the period of $f$ is $2\pi$ } \\&=\int_{-\pi}^\pi f(x)dx \end{align}

Do we need to prove the equality $\int_\pi^{\pi+a}f(x)dx=\int_{-\pi}^{-\pi+a}f(x+2\pi)dx$ ? If the answer is yes, how? Thanks!

Answer 1

\begin{align} \int_\pi^{\pi+a}f(x)dx &=\int_{\Bbb R}\chi_E(x)f(x)dx=\int_{\Bbb R}\chi_E(x+2\pi)f(x+2\pi)dx \\&=\int_{\Bbb R}\chi_{E-2\pi}(x)f(x)dx=\int_{-\pi}^{-\pi+a}f(x)dx \end{align}

Answer 2

$a$ need not be in $(0,2\pi]$ but this doesn't affect the proof. Yes, you have to justify your second step. Just make the substitution $x=y+2\pi$ for the justification.

Answer 3

There is a more general result, that follows from the translation invariance of Lebesgue measure, and makes precise the statement "$y=x−2π$ implies $dy=dx$ by differentiation":

Set $g:\mathbb R\to \mathbb R$ by $g(x)=x+a.$ Then, the image measure $mg^{-1}$ satisfies $mg^{-1}(E)=m(E)$ for an arbitrary measurable set $E$, and so

on the one hand,

$\int_Ef(x)d(mg^{-1})=\int_Ef(x)dm$

and on the other,

$\int_Ef(x)d(mg^{-1})=\int_{\mathbb R}\chi_E(x)f(x)d(mg^{-1})=\int_{\mathbb R}g(f(x)\chi_E(x))dm=\int_{\mathbb R}f(x+a)\chi_E(x+a))dm.$

Therefore,

$\int_Ef(x)dm=\int_{\mathbb R}f(x+a)\chi_E(x+a))dm$.

To recover your result, set $a=2\pi,\ E=(-\pi,\pi]$ and note that $f(x+2\pi)=f(x)$