Under what conditions can $L^p(X,\sigma,\mu)$ be an inner product space for $p\neq 2$? My measure theory book* claims that $L^p(X,\sigma,\mu)$ is an inner product space only for $p=2$, where it is implied that the inner product induces the $p$-norm on $L^p$. I have seen proofs of this statement for specific $L^p$ spaces, like $\ell^p(\mathbb N)$ (here) and $L^p[0,1]$ (here). 
However, I think this claim cannot be true in general, because of the following counterexample. Let $(X,\sigma)$ be any measurable space and let $\mu_0$ denote the trivial measure $\mu(A)=0$ for all $A\in\sigma$. Then $L^p(X,\sigma,\mu)$ is the Banach space $\{0\}$, which is trivially an inner product space.
Are there any other examples where $L^p(X,\sigma,\mu)$ is an inner product space (where the inner product induces the $p$-norm) with $p \neq 2$? If so, what are the strictest conditions on $(X,\sigma,\mu)$ under which $L^p(X,\sigma,\mu)$ is an inner product space only for $p=2$ ?

*René Schilling, Measures, Integrals, and Martingales (2nd Edition), p. 326.

An inner product space (over $\mathbb R$) satisfies the parallelogram law
$$ \|x + y\|^2 + \|x - y\|^2 = 2 \|x\|^2 + 2 \|y\|^2$$
Another case where all $L^p$ are inner product spaces is where there is only one nonempty member $A$ of the $\sigma$-algebra $\sigma$ such that $0 < \mu(A) <\infty$ (and thus $L^p(X,\sigma,\mu)$ is one-dimensional).  
However, if there exist disjoint $A, B \in \sigma$ such that $0 < \mu(A), \mu(B) < \infty$, let $x$ and $y$ be the indicator functions of $A$ and $B$ respectively.  Then
$ \|x + y\|_p = \|x - y\|_p = (\mu(A) + \mu(B))^{1/p}$ while
$\|x\|_p = \mu(A)^{1/p}$ and $\|y\|_p = \mu(B)^{1/p}$.  The parallelogram law for $x$ and $y$ reduces to
$$ (\mu(A)+\mu(B))^{2/p} = \mu(A)^{2/p} + \mu(B)^{2/p} $$
which is true only if $p=2$: for $p > 2$ the function $f(t) = t^{2/p}$ is strictly subadditive on $(0,\infty)$, while for $p < 2$ it is strictly superadditive.
