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 $\mathrm{\Omega}$, so that $\mu (\mathrm{\Omega})=1$. If $f$ is a real function in ${L}^{1}(\mu )$, if $a<f(x)<b$ for all $x\in \mathrm{\Omega}$, and if $\phi $ is convex on $(a,b)$ then

$\phi {\textstyle (}{\int}_{\mathrm{\Omega}}f\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu {\textstyle )}\le {\int}_{\mathrm{\Omega}}(\phi \circ f)\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu .$

After a few steps Rudin obtains the following inequality:

$\phi (f(x))-\phi (t)-\beta (f(x)-t)\ge 0$

for every $x\in \mathrm{\Omega}$. Here, $t={\int}_{\mathrm{\Omega}}f\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu \in (a,b)$ and $\beta $ is a real number. If $\phi \circ f\in {L}^{1}(\mu )$ then the inequality can be obtained by integrating both sides. However, I am not sure how to proceed in the case $\phi \circ f\notin {L}^{1}(\mu )$. Rudin states that in this case, the integral ${\int}_{\mathrm{\Omega}}(\phi \circ f)\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu $ is defined in the extended sense

${\int}_{\mathrm{\Omega}}(\phi \circ f)\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu ={\int}_{\mathrm{\Omega}}(\phi \circ f{)}^{+}\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu -{\int}_{\mathrm{\Omega}}(\phi \circ f{)}^{-}\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu $

where

${\int}_{\mathrm{\Omega}}(\phi \circ f{)}^{+}\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu =\mathrm{\infty}\text{and}{\int}_{\mathrm{\Omega}}(\phi \circ f{)}^{-}\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu \text{is finite.}$

I am unable to show the existence of ${\int}_{\mathrm{\Omega}}(\phi \circ f)\phantom{\rule{mediummathspace}{0ex}}\text{d}\mu $ as defined above.

Any help would be greatly appreciated. Thank you.