# Integration with simple $L^2$ functions

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 $$h\in L^2(\Omega,\Sigma,\mu)$$ ?

• For the first part, what can you say about $\int (f-g)h\ d\mu$? – Bungo Feb 14 at 3:31
• The second is near immediate from Hölder. – Sean Roberson 2 days ago

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.