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Can you tell me if this is correct? Thanks.

Claim: The following inner product is well-defined, i.e. finite for all $f,g \in L^2$: $ (f,g) = \int_X f \overline{g} d \mu$

Proof: We may assume $f,g$ are real-valued since we can otherwise write them as sums of real and imaginary part. We can also assume that they are $\geq 0$ since otherwise we can write $f = f^+ - f^-$ for $f^+ , f^- \geq 0$. By assumption, $f,g \geq 0$ are measurable hence there are sequences of simple functions $s_n \to f, t_n \to g$ converging pointwise and $s_n \leq s_{n+1}, t_n \leq t_{n+1}$ and hence $s_n t_n \to fg$ pointwise and $s_n t_n$ is non-decreasing. Simple functions form an inner product space with respect to $\langle .,.\rangle$ so Cauchy Schwarz holds. Then

$\begin{align} \int_X fg d \mu = \int_X \lim_{n \to \infty} s_n t_n d \mu & \stackrel{\text{monotone conv.}}{=}\lim_{n \to \infty} \int_X s_n t_n d \mu \\ & = \lim_{n \to \infty} \int_X |s_n| | t_n| d \mu \\ & \stackrel{\text{Cauchy Schwarz}}{\leq} \lim_{n \to \infty} \|s_n \|_2 \|t_n \|_2 \\ & \stackrel{\text{continuity of norm}}{=} \|f \|_2 \|g \|_2 < \infty \end{align}$

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    There is another point to be made: remember that elements of $L^2$ are not actually functions, but *equivalence classes* of functions. I would argue that you should also show that if $f\sim f'$ and $g\sim g'$, then $\langle f,g\rangle = \langle f',g'\rangle$.2012-07-16

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