If $h\in L^2(\Omega,\Sigma,\mu)$ is a simple function and $f=g\hspace{0.3cm} \mu-a.e.$ then $$\int fhd\mu=\int ghd\mu \qquad \text{and} \qquad \left|\int ghd\mu\right|\le\|f\|_{L^2}\|h\|_{L^2}$$ How can I prove that the above is true for all $h\in L^2(\Omega,\Sigma,\mu)$ ?

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    $\begingroup$ For the first part, what can you say about $\int (f-g)h\ d\mu$? $\endgroup$ – Bungo Feb 14 at 3:31
  • 1
    $\begingroup$ The second is near immediate from Hölder. $\endgroup$ – Sean Roberson Feb 14 at 4:35

Since $h\in L^2(\Omega,\Sigma,\mu)$ is a simple function, it is of the form $h(x) = \sum_{i=1}^n \alpha_i \chi_{E_i}(x)$, where $E_i \in \Sigma, \alpha_i \in \mathbb{R}$. Let $M:= \{x \in \Omega: g(x) \neq f(x) \}$. The set is measurable because $f$ and $g$ are measurable. We know that $\mu(M) = 0$. Set $M_i := M \cap E_i$. Then, $\int_{\Omega} fh d\mu = \int_{\Omega} f(x) \sum_{i=1}^n \alpha_i \chi_{E_i}(x) d\mu(x) = \sum_{i=1}^n \alpha_i \mu(E_i) \int_{E_i}f(x) d\mu(x) = \sum_{i=1}^n \alpha_i \mu(E_i) (\int_{E_i}f(x) d\mu(x) + \int_{M_i}g(x) d\mu(x)) = \sum_{i=1}^n \alpha_i \mu(E_i) \int_{E_i}g(x) d\mu(x) = \int_{\Omega} gh d\mu$

To answer your second equation: $| \int gh d\mu | = | \int fh d\mu | = \|fh\|_1 \le \|f\|_2 \|h\|_2$ which is the Hölder's inequality.


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