# Is my interpretation of why $\|\cdot\|_p$ becomes a norm on $L^p$ correct?

I'm new to measure theory, $$L^p$$ spaces, etc. I was wondering if my interpretation of why $$\|\cdot\|_p$$ becomes a norm on $$L^p$$ is correct.

Consider a measure space $$(X,\mathcal{A},\mu)$$, and $$\mathcal{L}^p = \{f : X \to \mathbb{C}, f \ \text{is measurable}, \|f\|_p < \infty \}$$, where $$\|f\|_p = \left( \int \lvert f \rvert^p \mathrm{d}\mu \right)^{1/p}$$, $$p \in [1,\infty)$$ is a seminorm on $$\mathcal{L}^p$$.

Define an equivalence relation on $$\mathcal{L}^p$$ as $$f \sim g \iff f = g, \ \mu \ \text{a.e.}$$

Let $$L^p$$ be the set of equivalence classes. Then $$(L^p, \|\cdot\|_p)$$ becomes a normed space.

My understanding is that $$\|\cdot\|_p$$ starts out as a seminorm because it can be $$0$$ if $$f$$ is $$0 \ \mu$$-a.e. For $$\|\cdot\|_p$$ to be a norm, $$\|f\|_p = 0 \iff f = 0$$. This property is achieved when restricting $$\|\cdot\|_p$$ to $$L^p$$, since if a function $$f$$ is $$0 \ \mu$$-a.e., it is in the same equivalence class as $$0$$, hence $$f$$ and $$0$$ are "treated as the same thing". Is this a correct interpretation?

• Yes, it is correct. Apr 17, 2020 at 0:11