The decimal expansion $T:([0,1],\mathcal{B}_{[0,1]},\lambda)\longrightarrow ([0,1],\mathcal{B}_{[0,1]},\lambda)$ is defined to be $$T(x)=10x\ \ \text{mod}\ \ 1,$$ where $\lambda$ is Lebesgue measure.
I know how to prove that $T$ preserves $\lambda$. And I am now trying to prove that $T$ is ergodic. To avoid complicated set manipulation, I want to do this using Fourier series.
Suppose there exists $f\in L^{2}([0,1],\mathcal{B}_{[0,1]},\lambda)$ such that $f\circ T=f$ $\lambda-$almost surely, I want to show $f$ is constant $\lambda-$almost surely.
Note that if $f\circ T=f$ $\lambda-$a.s. then $f\circ T^{p}=f$ $\lambda-a.s.$ for all $p\in\mathbb{Z}_{>0}$. Now, consider the Fourier series of $f$: $$f\sim \sum_{n=-\infty}^{\infty}c_{n}e^{2\pi nx}.$$ Then, $f\circ T^{p}$ has the Fourier series $$f\circ T^{p}\sim \sum_{n=\infty}^{\infty}c_{n}e^{2\pi n 10^{p}x}.$$ This two Fourier series should be the same, so comparing coefficients we see $$c_{10^{p}n}=c_{n}.$$ Suppose $n\neq 0$, then $10^{p}n\longrightarrow\infty$, as $n\rightarrow\infty$. It then follows from Riemann-Lebesgue lemma that $$c_{n}=c_{10^{p}n}\longrightarrow0,\ \text{as}\ n\rightarrow\infty.$$
Then I got stuck. How could I conclude from here that $c_{n}=0$ for all $n\neq 0$? Thank you so much!
Use the Plancherel identity: since $f \in L^2$, we must have $$ \infty > \int_{0}^{1} |f(x)|^2 \,dx = \sum_{n} |c_n|^2 $$ But you've shown that $c_n = c_{10n} = c_{100n} = \dots$, so if any $c_n$ is nonzero, then the right hand side is infinite, contradiction.