Complete function spaces and inner products

Question

I am reading the technical report by Leo Breimann entitled "Some Infinity Theory for Predictor Ensembles".

On page 4 he has the following:

Definition 1 Let $L_2(P)$ be the space of functions on $\mathbb{R}^D$ that are square - integrable wrt $P(dx)$. A set of functions $F$ in $L_2(P)$ will be called complete if the $L_2$-closure of all finite linear combinations of functions in $F$ by $L_2(F)$, equals $L_2(P)$.

And then the following property:

Property 1 A set of functions, $F$, is complete iff there is no non-zero fucntion $g$ in $L_2(P)$ s.t. $\langle g,f \rangle = 0 \forall f\in F$, where $\langle g,f \rangle = \int f(x)g(x)P(dx).$

I do not understand how Property 1 works. As far as my intuition goes, a zero inner product implies orthogonality. But any vector/function space should be able to be decomposed into an orthogonal set of basis vectors.

Thus property 1 as I understand seems to suggest that there cannot exist a function $g$ which makes the inner product 0,and so by implication there cannot exist an orthogonal set of basis vectors on a complete function space?

So property 1 seems to link completeness and orthogonality in a way which I can't fully grasp. I was hoping someone here could help clarify this understanding for me.

Link to paper: https://statistics.berkeley.edu/tech-reports/579

Answer 1

As the comment has pointed out

You are missing the "for all f" part of the property. It can exist functions orthogonal to g, but not all

Thus there is allowed to exist a "zeroing" function for some $f$ but it cannot be true "for all" $f$, otherwise there will be "holes" in the space naturally.