How do functions form a Hilbert space I'm trying to wrap my head around function spaces. I get that you can define the inner product as the integral the multiplication of two functions over the entire domain because it satisfies the properties of the inner product.
What I don't understand is what makes this a complete space? I don't understand the concept how a complete space works with functions.
Any insight would be appreciated.
 A: What does it mean for a metric space $(X,d)$ to be complete? It means that if $\{x_n\}_{n=1}^\infty$ is Cauchy sequence then there's some $x \in X$ such that 
$$\lim_{n\rightarrow \infty} x_n=x.$$
So if we have a measure space $X$ and we talk about the space of functions (really equivalence classes) $L^2(X)$ of functions that satisfy
$$\int_X f^2< \infty.$$
We can define a norm on them by 
$$||f||=\sqrt{\int_x f^2},$$
the fact that this is a norm is Minkowski's inequality. Now what does it mean to say that $L_2(X)$ is complete? It means that if we have some Cauchy sequence of functions $\{f_n\}_{n=1}^\infty$ then the $f_n$ converge to a function $f \in L^2(X)$. As Kevin mentions this is the Riesz-Fischer Theorem. It makes use of the fact that in normed vector spaces we don't have to work quite as hard as in metric spaces. It's sufficient to show that if we have a sequence $\{f_n\}_{n=1}^\infty$ such that 
$$\sum_{n=1}^\infty ||f_n|| < \infty$$
then the sum converges to some $f$. Which in our case a simple consequence of the monotone and dominated convergence theorems. 
A: To have a complete space, you need a norm (at least a metric.) On the Hilbert space of functions, $L^2(X)$ for $X$ some measure space, we define a norm
$$||f||=\sqrt{\int f^2 d\mu}$$
That $L^2(X)$ is complete is the Riesz-Fischer theorem.
http://en.wikipedia.org/wiki/Riesz-Fischer_theorem
