# Question regarding a step in Rudin's proof of the Jensen's inequality

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.

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 $$\int (\phi \circ f)^{-} d\mu \leq \int g^{-} d\mu <\infty$$].