I am reading Rudin's RCA and have a question regarding the proof of Jensen's inequality.
Let $\mu$ be a positive measure on a $\sigma$-algbebra $\mathfrak{M}$ in a set $\Omega$, so that $\mu(\Omega) = 1$. If $f$ is a real function in $L^{1}(\mu)$, if $a < f(x) < b$ for all $x\in\Omega$, and if $\varphi$ is convex on $(a, b)$ then $$\varphi\Big(\int_{\Omega}f\:\text{d}\mu\Big)\leq \int_{\Omega}(\varphi\circ f)\:\text{d}\mu.$$
After a few steps Rudin obtains the following inequality: $$\varphi(f(x))-\varphi(t)-\beta(f(x)-t)\geq 0$$ for every $x\in \Omega$. Here, $t = \int_{\Omega}f\:\text{d}\mu\in(a, b)$ and $\beta$ is a real number. If $\varphi\circ f\in L^{1}(\mu)$ then the inequality can be obtained by integrating both sides. However, I am not sure how to proceed in the case $\varphi\circ f\notin L^{1}(\mu)$. Rudin states that in this case, the integral $\int_{\Omega}(\varphi\circ f)\:\text{d}\mu$ is defined in the extended sense $$\int_{\Omega}(\varphi\circ f)\:\text{d}\mu = \int_{\Omega}(\varphi\circ f)^{+}\:\text{d}\mu-\int_{\Omega}(\varphi\circ f)^{-}\:\text{d}\mu$$ where $$\int_{\Omega}(\varphi\circ f)^{+}\:\text{d}\mu = \infty\text{ and }\int_{\Omega}(\varphi\circ f)^{-}\:\text{d}\mu\text{ is finite.}$$ I am unable to show the existence of $\int_{\Omega}(\varphi\circ f)\:\text{d}\mu$ as defined above.
Any help would be greatly appreciated. Thank you.
Best Answer
Let $g(x)=\phi (t)+\beta (f(x)-t)$. Then $(\phi \circ f) (x)\geq g(x)$ for all $x$ and $g$ is integrable. Writing $\phi \circ f =(\phi \circ f-g)+g$ and noting the first term is non-negative and the second term is integrable we see that $\int (\phi \circ f) d\mu$ exists.
[ $(\phi \circ f) \wedge 0 \geq g \wedge 0$ so $(\phi \circ f)^{-} \leq g^{-}$. Hence $\int (\phi \circ f)^{-} d\mu \leq \int g^{-} d\mu <\infty$].