[This is an exercise of Measure and Integration. I am repeating this in my vacation.]
Define $f:[a,b] \rightarrow R$ a function such that $f(x) \geq 0$ over $[a,b]$ and f is R-Integrable in [a,b]. Show that $\sqrt{f}$ is R-Integrable in [a,b].
MY ATTEMPT
I prove that:
$f(x): [a,b] \rightarrow [-M,M]$ is R-Integrable in $[a,b]$ and $\varphi$ is a continuous funcion (*) such that $\varphi:[-M,M] \rightarrow R$ so $\varphi \circ f$ is a R-Integrable function in $[a,b]$
So I conclude that if $\varphi: [\inf\{f\};\sup\{f\}] \rightarrow R$ such that $\varphi=\sqrt{x}$ we have the result, since $\sqrt{x}$ is a continuous fuction.
But, I'm not satisfied by this.
MY QUESTIONS
1- how can I ensure the existence of $\inf$ or $\sup$ of $f$ (f is not necessarily continuous); (solved by @copper.hat)
2- how else I can prove this exercise? (my questioning came because the order of the course we learned that the composite of a continuous function with a R-Integrable is R-Integrable after this)
EXTRA QUESTION (about (*))
We can "improve" this part assuming only $\varphi$ is R-Integrable?
Edit:
My answer for the extra question:
Let:
$$
f(x)= \left \{ \begin{array}{ll}
0 & \textrm{ se } x \not\in Q \\
1/q & \textrm{ se } x=p/q, (p,q)=1, 0 <x \leq 1\\
1 & \textrm{ se } x=0 \\
\end{array}
\right .
$$
$$
g(x)= \left \{ \begin{array}{ll}
0 & \textrm{ se } x =0 \\
1 & \textrm{ se } 0 <x \leq 1\\
\end{array}
\right .
$$
$f,g$ are R Integrables but the $g\circ f$ is the Dirichlet Function.
Best Answer
Here's some food for thought: The square root function is Lipshitz when restricted to values $\ge\delta$, where $\delta>0$. This follows from $$\sqrt{u}-\sqrt{v}=\frac{u-v}{\sqrt{u}+\sqrt{v}},$$ say.
Secondly, the parts where $f(x)<\delta$, or rather where some step function $\ge f$ is smaller than $\delta$, are of negligible consequence.
These two facts should be enough to let you work out a solution.