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As usual denote $L^p$ the quotient space where two integrable functions are identified if they are equal almost everywhere. So I'm using the definition written here:

http://en.wikipedia.org/wiki/Lp_space

Then we have the following result: for each $p \geq 1$ we have $L^{\infty}(X) \subseteq L^{p}(X)$ where X is a finite measure space and $L^{\infty}$ denotes the set of all essentially bounded functions endowed with the $||f||_{\infty}$ pseudonorm.

So I took $f \in L^{\infty}(X)$ then by definition there is some bounded function $g$ such that $g=f$ a.e. But then $f=g$. So:

$\int |f|^{p} = \int |g|^{p} \leq \int (||g||_{\sup})^{p} < \infty$.

Questions: Is the above correct? Why do we need $p \geq 1$. Why wouldn't p>0 work? Is it because we need $p \geq 1$ in the case q is not $\infty$ or where exactly?

Thank you.

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    @user:Technically, $g=f$ a.e. means that *the equivalence classes* of $f$ and $g$ are equal, rather than $f$ and $g$ necessarily being equal. We usually abuse notation in $L^p$ spaces and use the function to denote its class, but here it might be somewhat confusing to say "$f=g$ a.e., so $f=g$".2010-11-10

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If p<1, then the integral $(\int |f|^p d \mu)^{\frac{1}{p}}$ no longer defines a norm. That is why we need $p \geq 1$ typically. However, for the conclusion you want to draw, I guess it will still work if $p \geq 0$.

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    @user: $\|[g]\|_\sup$ doesn't really make sense, strictly speaking. Also, using $[f]$ and $[g]$ in the integrals isn't necessary and possibly a little confusing. When it comes down it, we're integrating functions, we just also keep in mind that it doesn't matter which representative of the equivalence class we integrate. The way you wrote the integrals in your question is actually good. Arturo's point is only about saying $f=g$; it is true that they represent the same element of $L^\infty$, but since you're dealing explicitly with representatives it's best not to abuse notation too much.2010-11-10