Why do we need $|\cdot|$ in this statement: $f \in L^p \iff |f|^p \in L^1$?

Question

I am somewhat confused about the following statement: $$f \in L^p \iff |f|^p \in L^1$$

Why do we need the $|\cdot|$ if we have it in $$ ||f ||_p := \bigg(\int |f| ^p\bigg)^{\frac 1 p}$$ anyway?

What breaks down if we drop $|\cdot|$ and write $$f \in L^p \iff f^p \in L^1$$

Answer 1

The absolute value is not necessary for the definition of $L^p$ as elements such that $f^p$ is integrable. The reason is that a function is Lebesgue integrable iff its absolute value is integrable.

Now, when you are going to define $||f||_p$ you want it to be a norm. If you don't put the absolute value in the definition you are not going to get a norm.

Answer 2

If the statement that you quoted came from a context where real-valued function spaces were under consideration, then the answer is given by GEdgar's comment: If $f$ has negative values and $p$ is not an integer, then $f^p$ isn't available as a real-valued function. If, on the other hand, the function spaces consist of complex-valued functions, then $f^p$ might not be single-valued. That's not a major problem, because you just have to choose a single-valued determination of it, and the quoted statement without the absolute value will be true for any reasonable choice. But the author would have to say something about making that choice, and it's quicker just to avoid the issue by using the absolute value, especially since, as Franklin.vp pointed out, the absolute value is what's relevant to the norm.