I understood the whole proof, but I can't understand the underlined part. Rudin says that if $\varphi \circ f$ is not in $L^1(\mu)$ then the integral of $\varphi \circ f$ is $+ \infty$ in the sense of Section 1.31 as the proof shows. But I can't see how the proof implies this part.
The integral in Section 1.31 is defined as follows: For a measurable function $f: X\to [-\infty,\infty]$, we define $\int _X f d\mu= \int _X f^+ d\mu-\int _X f^- d\mu$, provided that at least one of the integrals on the right-hand-side is finite, where $f^+, f^-$ are the positive and the negative parts of $f$, respectively.
Thanks in advance.
The proof shows that $g(x)=\varphi(f(x))-\varphi(t)-\beta(f(x)-t)$ is a positive measurable function on $\Omega$. Hence, $\varphi\circ f=g+\varphi(t)+\beta(f-t)$ is the sum of a positive measurable function $g$ and an element of $L^1(\mu)$, hence the integral of $\varphi\circ f$ exists in the extended sense. Its integral, if not finite, can only be $+\infty$ as again, $\varphi\circ f$ is the sum of a positive measurable function and a function in $L^1(\mu)$.