The theory of integration of real functions is (as far as I know) usually extended to the complex case as follows:
Let $X$ be a set. Given a function $f:X\to\mathbb C$, define $\Re(f),\Im(f):X\to\mathbb R$ as $\Re(f)(x)=\Re(f(x))$ and $\Im(f)(x)=\Im(f(x))$ for every $x\in X$, where $\Re$ and $\Im$ denote the real part and the imaginary part respectively. Then, it clearly is the case that $f=\Re(f)+i\Im(f)$.
Now, if $(X,\mathscr A,\mu)$ is a measure space, we say that $f:X\to\mathbb C$ is integrable if and only if $\Re(f)$ and $\Im(f)$ are integrable, and we define \begin{align*} \int f~d\mu=\int\Re(f)~d\mu+i\int\Im(f)~d\mu. \end{align*} This implies in particular that \begin{align*} L^1(X,\mathscr A,\mu,\mathbb C)=L^1(X,\mathscr A,\mu,\mathbb R)+i\cdot L^1(X,\mathscr A,\mu,\mathbb R),\tag{1} \end{align*} where $L^p(X,\mathscr A,\mu,\cdot)$ denotes the usual $L^p$ space.
Question. Equation (1) raises for me an interesting question: is it true that, for any $1\leq p<\infty$, \begin{align*} L^p(X,\mathscr A,\mu,\mathbb C)=L^p(X,\mathscr A,\mu,\mathbb R)+i\cdot L^p(X,\mathscr A,\mu,\mathbb R) \end{align*} holds? If not, for which $p$ does it hold and for which $p$ does it fail?
Attempted solution: By definition, $f\in L^p(X,\mathscr A,\mu,\mathbb C)$ means that $f^p$ is integrable. Now, working with $\Big(\Re(f)+i\Im(f)\Big)^p$ seems very cumbersome to me, so instead we could define functions $r:X\to[0,\infty)$ and $\theta:X\to[0,2\pi)$ such that $f(x)=r(x)\exp(i\theta(x))$. In this case, we have that \begin{align*} f=r\cos(\theta)+ir\sin(\theta). \end{align*} and \begin{align*} f^p=r^p\cos(p\theta)+ir^p\sin(p\theta). \end{align*} However, I can't seem to get any further with this.