Riesz Representation Theorem: what is meant by extension by continuity

Question

I'm reading through these notes, but I'm a little concerned about this sentence in the proof of one of the theorems; "Since simple functions are dense in $L^p$ and $g\mapsto \int fg \ d\mu$ is a continuous functional on $L^p$ when $f\in L^{p'}$, it follows $F(g)=\int fg\ d\mu$ for every $g\in L^p(X)$." Here $F$ is the bounded linear functional on $L^p$ and assume $\mu(X)<+\infty$.

I'm able to show that $L_f:L^p\to\mathbb{R}$ defined by $L_f(g)=\int gf\ d\mu$ is a bounded linear functional, hence continuous, for $f\in L^{p'}$. However, this extension by continuity is a little vague to me. Since $\mu(X)<\infty$, simple functions of the form $$\phi=\sum_{i=1}^nc_i\chi_{A_i}$$ are dense in $L^p(X)$. Therefore, if $g\in L^p(X)$, then there is a sequence $\{\phi_n\}$ such that $\lim_n\|\phi_n-g\|_p=0$. However, I'm not sure how this would help to show that for any $g\in L^p$, $F(g)=L_f(g)$ using continuity.

Most resources I've looked at give this vague (at least to me it's vague) statement "now extend by continuity" or something similar. What is the usual argument when one encounters such a statement? I know the idea is that if a dense subset of $L^p$ has some property, then we can "extend by continuity", but I'm not sure what this means in the form of a proof. Any help would be greatly appreciated.

Answer 1

The sentence you say you're concerned about has more or less nothing to do with "extension by continuity". A careful justification (assuming my guesses about the meaning of the notation are correct):

Say $S\subset L^p$ is the set of simple functions. Define $\Lambda:L^p\to\Bbb R$ by $$\Lambda g=Fg-\int fg.$$Since $\Lambda$ is continuous, $\Lambda g=0$ for all $g\in S$ and $S$ is dense in $L^p$ it follows that $\Lambda g=0$ for all $g\in L^p$.

This uses the BFADS:

Basic Fact About Dense Subsets Suppose $X$ is a metric space, $D\subset X$ is dense, $f:X\to\Bbb R$ is continuous, and $f(x)=0$ for all $x\in D$. Then $f(x)=0$ for all $x\in X$.

Proof: Suppose $x\in X$. Let $\epsilon>0$. Choose $\delta>0$ so that $d(x,y)<\delta$ implies $|f(y)-f(x)|<\epsilon$. Since $D$ is dense, there exists $y\in D$ with $d(y,x)<\delta$. Hence $|f(x)|=|f(x)-f(y)|<\epsilon$.

And so $|f(x)|<\epsilon$ for every $\epsilon>0$, hence $f(x)=0$.

Answer 2

Try thinking about it this way: your functional is a linear continuous function from the step functions to $\mathbb{R}$. Because of the linearity the continuity of the functional is equivalent to its uniform continuity. Now it is well known that a uniformly continuous function can be uniquely extended (as a uniformly continuous function) to the completion of its domain (here the completion is $L^p$, since the step functions are dense). That is what is meant by that phrase.