OK.  Perhaps a reason $M^p$ is not often studied is: it is not even a  vector space.
(using the original defintion with lim not limsup.)  

Define functions $f$ and $g$ as follows:  
$f(x)=0$ if $x<1$,  
$f(x)=1$ if $x \ge 1$ and $\{x\}< 1/2$; here, $\{x\} = x-\lfloor x\rfloor$ is the fractional part  
$f(x)=-1$ if $x \ge 1$ and $\{x\} \ge  1/2$.  

$g(x)=0$ if $x<1$,  
$g(x)=f(x)$ if $x \ge 1$ and $\lfloor \log_2(x)\rfloor$ is even,  
$g(x)=-f(x)$ if $x \ge 1$ and $\lfloor \log_2(x)\rfloor$ is odd.  

Some graphs:

$f(x)$  
![f][1]  

$g(x)$  
![g][2]  

$f(x)+g(x)$  
![f+g][3]  

But note:
$$
\lim_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)|^2\right)^{1/2} =
\lim_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |g(x)|^2\right)^{1/2} = \frac{1}{\sqrt{2}}
$$
both exist, while
$$
\lim_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2}
$$
does not exist.  In fact (do some computations):
$$
\limsup_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2}
=\frac{2}{\sqrt{3}},
$$  

$$
\liminf_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2}
=\frac{\sqrt{2}}{\sqrt{3}}
$$


**added**  
With the "limsup" definition, as suggested by Jean Van Schaftingen, we contradict the parallelogram law, since
$$
\|f\|_{M^2} = \|g\|_{M^2} = \frac{1}{\sqrt{2}},\qquad
\|f+g\|_{M^2} = \|f-g\|_{M^2} = \frac{2}{\sqrt{3}}
$$


  [1]: https://i.sstatic.net/8U20W.jpg
  [2]: https://i.sstatic.net/eTDbG.jpg
  [3]: https://i.sstatic.net/YTLZD.jpg