As far as I understand, $\mu$ is a Borel probability measure on teh unit circle $\mathbb{T}$. Then $$ \frac{1}{2N+1}\sum\limits_{n=-N}^N|\hat\mu(n)|^2=\int\int\frac1{2N+1}\sum_{n=-N}^N (x/y)^nd\mu(x)d\mu(y). $$ The integrand $$ \frac1{2N+1}\sum_{n=-N}^N (x/y)^n $$ has absolute value at most 1 and converges pointwise to 1 for $x=y$ and to $0$ for $x\ne y$ (the sum is a geometric progression, you may sum it up to see that it is bounded when $x\ne y$). Thus by Dominated Convergence theorem the limit is the $\mu\times \mu$-measure of the diagonal $\{x=y\}$, which is exactly LHS.

As far as I understand, $\mu$ is a Borel probability measure on the unit circle $\mathbb{T}$. Then $$ \frac{1}{2N+1}\sum\limits_{n=-N}^N|\hat\mu(n)|^2=\int\int\frac1{2N+1}\sum_{n=-N}^N (x/y)^nd\mu(x)d\mu(y). $$ The integrand $$ \frac1{2N+1}\sum_{n=-N}^N (x/y)^n $$ has absolute value at most 1 and converges pointwise to 1 for $x=y$ and to $0$ for $x\ne y$ (the sum is a geometric progression; you may sum it up to see that it is bounded when $x\ne y$). Thus, by the Dominated Convergence Theorem, the limit is the $\mu\times \mu$-measure of the diagonal $\{x=y\}$, which is exactly the LHS.