Let $ \left\{ X\left(t\right)\right\} _{t\in R} $ be a WSS process.
There is a number $T\in \mathbb{R} $ such that $R_X(0)=R_X(T) $ where $R_X$ is the autocorrelation function. Prove that $R_X$ is a periodic function with period $T$.
My attempt: $$ \require{cancel} R_{X}\left(0\right)=R_{X}\left(t_{i},t_{i}\right)=R_{X}\left(t_{i},t_{i}+T\right)\tag{Given} \\ $$ Now we can multiply both sides by $R_x(t_i)$ $$ R_{X}\left(t_{i},t_{i}\right)\cdot R_{X}\left(t_{i}\right)=R_{X}\left(t_{i},t_{i}+T\right)R_{X}\left(t_{i}\right)\Rightarrow$$ $$ \\ R_{X}\left(t_{i},t_{i}\right)\cdot R_{X}\left(0,t_{i}\right)=R_{X}\left(t_{i},t_{i}+T\right)\cdot R_{X}\left(0,t_{i}\right)\\ \int f_{X\left(t_{i}\right)}f_{X\left(t_{i}\right)}X\left(t_{i}\right)X\left(t_{i}\right)\int f_{X\left(0\right)}f_{X\left(t_{i}\right)}X\left(0\right)X\left(t_{i}\right)=\int f_{X\left(t_{i}\right)}f_{X\left(t_{i}+T\right)}X\left(t_{i}\right)X\left(t_{i}+T\right)\int f_{X\left(0\right)}f_{X\left(t_{i}\right)}X\left(0\right)X\left(t_{i}\right)\\ $$ Now assuming we can rearenge the 2 integrals on the right we can get: $$ \cancel{\int f_{X\left(t_{i}\right)}f_{X\left(t_{i}\right)}X\left(t_{i}\right)X\left(t_{i}\right)}\int f_{X\left(0\right)}f_{X\left(t_{i}\right)}X\left(0\right)X\left(t_{i}\right)=\int f_{X\left(0\right)}f_{X\left(t_{i}+T\right)}X\left(t_{i}+T\right)X\left(0\right)\cancel{\int f_{X\left(t_{i}\right)}X\left(t_{i}\right)f_{X\left(t_{i}\right)}X\left(t_{i}\right)} \Rightarrow R_{X}\left(t_{i}\right)=R_{X}\left(t_{i}+T\right)$$
Is this solution valid, assuming interchanging elements in two integrals are allowed?