Proof for the normalized average power of a periodic signal I'm trying to understand why the power of a periodic signal, with fundamental period $T_o$ is defined as
$$P = \frac{1}{T_0}\int_0^{T_o}|x(t)|^2 dt\tag{1}\label{1}$$
I know the definition of power is
$$\lim_{T\to \infty} \frac{1}{T} \int_{-T/2}^{T/2}|x(t)|^2 dt \tag{2}\label{2}  $$
and I know the key is to take the integral multiple of the power over the normalized frequency range or 
$$k\int_0^{T_o}|x(t)|^2 dt\tag{3}\label{3}$$
but I'm missing the steps between these step 3 and step 2. Any help would be greatly appreciated. 
 A: Let $M = \int_{-T_0}^{T_0} |x(t)|^2\,dt$ Note that for any integer $k > 0$ and any $0< \tau < T_0$, we can use the periodicity of $x$ to deduce that
$$
\frac 1{2kT_0 + \tau}\int_{-kT_0 - \tau}^{kT_0 + \tau} |x(t)|^2 \,dt = \\
\frac 1{2kT_0 + \tau}\left(\int_{-kT_0}^{kT_0} |x(t)|^2 \,dt + \int_{-\tau}^{0} |x(t)|^2 \,dt + \int_0^\tau |x(t)|^2\,dt\right) \leq\\
\frac 1{2kT_0 + \tau}\left(\int_{-kT_0}^{kT_0} |x(t)|^2 \,dt  + M\right) = \\
\frac 1{2kT_0 + \tau}\int_{-kT_0}^{kT_0} |x(t)|^2 \,dt + \frac{M}{2kT_0} \leq \\
\frac 1{2kT_0}\int_{-kT_0}^{kT_0} |x(t)|^2 \,dt + \underbrace{\frac{M}{2kT_0}}_{\text{goes to zero}}
$$
From the other side, 
$$
\frac 1{2kT_0 + \tau}\int_{-kT_0 - \tau}^{kT_0 + \tau} |x(t)|^2 \,dt = \\
\frac 1{2kT_0 + \tau}\left(\int_{-kT_0}^{kT_0} |x(t)|^2 \,dt + \int_{-\tau}^{0} |x(t)|^2 \,dt + \int_0^\tau |x(t)|^2\,dt\right) \geq\\
\frac 1{2kT_0 + T_0} \int_{-kT_0}^{kT_0} |x(t)|^2 \,dt = \\
\underbrace{\frac{2kT_0}{{2kT_0 + T_0}}}_{\text{goes to 1}}\frac 1{2kT_0} \int_{-kT_0}^{kT_0} |x(t)|^2 \,dt
$$
