I am currently reading Big Rudin and I have this theorem that I've been struggling with for some time now.
Usually, when reading, I either try to get a concrete intuition behind the idea, visualise the theorem and its proof in my head, or at least connect it to something I already know, as well as try to connect the proof to other proofs I know.
However, upon this theorem, I truly stumbled. As Rudin notes in the beginning of the chapter, one can see a little connection between the measure and linear functionals as $b-a$ can be approximated by $\Lambda f$ where $\Lambda f=\int_a^b fdx$ and $f$ is a continuous function on $[a;b]$ with range lying in $[0;1]$. However, this honestly felt like "cheating".
As to why this is cheating, it seems so to me for various reasons. First of all, he gave the most trivial of examples while one needs a general idea. However, the real reason is the fact that the linear function he uses is actually intimately related to measure by itself, without any need for a representation theorem. The ideas of measure and integrations are intertwined and thus stating that we can find a measure that represents an integral (especially since it's only on an interval) seems useless and uninformative.
Is there a way to intuitively understand why such a representation is possible or is this theorem too complex for any intuition?
Thanks in advance.
So the statement is that every continuous linear functional on $C[0,1]$ is given by integration against some Borel measure. First, the fact that Borel measures give continuous linear functionals is easy to prove, using pretty much whatever measure theoretic convergence theorem you want. That may seem obvious, but it is important for intuition.
To go the other way, recall that measuring a set is the same as integrating the indicator function of that set. So if $F$ is a continuous linear functional and $f_n$ converges to $1_A$ in some appropriate sense, then $\lim_{n \to \infty} F(f_n)$ is the natural candidate for $\mu(A)$. Thus you just need to construct such an approximating sequence. Can you do that if $A$ is assumed open? How can you extend that case to the general one?