I see in many real analysis books, for example the one I'm going to say, the author first proved the Integration by Substitution Indefinite Integral Version $\int f(g(x))g'(x)dx=F(g(x))+C$, which the proof is easy. But the next thing let me feel curious, why did he then proved the Integration by Substitution "Definite Integral" Version(see the figure) with so lengthy manner? Having know that $\int f(g(x))g'(x)dx=F(g(x))+C$, and combine it with the Fundamental Theorem of Calculus, the definite integral $\int_a^b f(g(x))g'(x)dx$ is simply the difference of its antiderivative $F(g(x))$ at two points $a$ and $b$, $[F(g(x))]^b_a$. Why we need a relatively lengthier proof here?
The Fundamental Theorem of Calculus says that if $F'(t) = f(t)$ and $f$ is integrable on $[a,b]$ then
$$ \int_a^b f(t) \,dt = F(b) - F(a). $$
Just knowing that the derivative of $F(g(x))$ is $f(g(x))g'(x)$ is not enough. You need to also know that $f(g(x))g'(x)$ is integrable. Well, when $g'(x)$ is continuous then $f(g(x))g'(x)$ is continuous and hence integrable, but we don't know that $g'(x)$ is continuous, only that it's integrable. So we need a different argument that shows that $f(g(x))g'(x)$ is integrable. This is easy enough to do and is what is done in the first paragraph.
The second paragraph is verifying that if $F$ is an antiderivative of $f$ then $F \circ g$ is an antiderivative of $(f \circ g)g'$. That's covered in the next 5 sentences (first say why $F$ exists, then say that $F' = f$ then say that $(F \circ g)' = (f \circ g)g'$).
The next line, which reads
Because $f$ is continuous then $f$ is integrable by Theorem 5.4.11, and we also know that $f$ has an antiderivative, which is $F$.
I'm not sure is strictly necessary, given that that is covered in the first two sentences of the second paragraph. If I had to guess, the author is trying to remind you of the conditions for Exercise 5.6.6.
The last sentence ties everything together.