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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:

$f(x)$

$g(x)$

$f(x)+g(x)$

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

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

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:

$f(x)$

$g(x)$

$f(x)+g(x)$

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 $$ \|f\|_{M^2} = \|g\|_{M^2} = \frac{1}{\sqrt{2}},\qquad \|f+g\|_{M^2} = \|f-g\|_{M^2} = \frac{2}{\sqrt{3}} $$

OK. Perhaps a reason $M^p$ is not often studied is: it is not even a vector space. 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:

$f(x)$

$g(x)$

$f(x)+g(x)$

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

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

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:

$f(x)$

$g(x)$

$f(x)+g(x)$

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

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

OK. Perhaps a reason $M^p$ is not often studied is: it is not even a vector space. 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:

$f(x)$

$g(x)$

$f(x)+g(x)$

But note: (continued)

OK. Perhaps a reason $M^p$ is not often studied is: it is not even a vector space. 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:

$f(x)$

$g(x)$

$f(x)+g(x)$

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

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