Here are definition from wiki:
inner product for Hilbert space, two square-integrable real-valued functions $f$ and $g$ on an interval $[a, b]$ $$\langle f,g\rangle=\int^a_b f(x)g(x)dx$$
inner product for Lebesgue space, consider $L_2(\Omega, \mu)$
$$\langle f,g\rangle=\int_{\Omega} f(x)g(x)d\mu(x)$$
Since $L_2$ is also Hilbert space so they should be the same, is $dx$ also a measure?
They are only differences in the domain wrt different $dx$ and $d\mu(x)$, and I thought it is change of variable problem, below is my motivation for this question push-froward measure and change of variable
I am so confused about those, please help me to clarify.
Extra thoughts
From Stein's real analysis I found: $$\|f\|_{L^2(\mathbb{R}^d)}^2=\int_{\mathbb{R}^d}|f(x)|^2dx$$
$$\|f\|_{L^2(X,\mu)}^2=\int_{X}|f(x)|^2d\mu$$
I guess they are really the same?
- In $\mathbb{R}^d$ so the set is Riemann integrable, $dx$ itself read as already a measure. In contrast to $L^2(X,\mu)$ we need a Lebesgue measure.
If the above reasoning is correct, what guarantee $\mathbb{R}^d$ is Riemann integrable, I mean rational indication function is not Riemann integrable.
Please help me to clarify my thoughts
From the viewpoint of Lebesgue theory, $dx$ as you are used to it is an abbreviation for $dm(x)$ where $m$ is the Lebesgue measure. Lebesgue theory also lets you integrate over other sets as long as you are considering real-, complex-, vector-, or even Banach space-valued functions on a measure space.