Define $h(x)=min(f(x),g(x))$ and $l(x)=\int_{a}^{x}f(t)dt$, then which of the following is/are true:
(a) $h$ need not be Riemann integrable but $l$ always is.
(b) $l$ need not be Riemann integrable but $h$ always is.
(c) $h,l$ are Riemann integrable, always.
(d) Whenever $h$ and $l$ are Riemann integrable, $\int_{a}^{x}h(t)dt ≤l(x)$ for all $x$ in $(a,b)$
Now it can be seen that $l'(x)=f(x)$ and since $f$ is Riemann integrable, it makes $l'(x)$ Riemann integrable and hence $l(x)$ integrable (idk if I'm correct in assuming so). I google $f$ and $f'$ relation in Riemann sum but couldn't find much.
I can't start analysing $h(x)$ since I'm not able to perceive how the minimum function changes the lower and upper Riemann Sums. Any help, mathematicians? I feel like a noobie around here sometimes!
Note that $l(x)$ is a continuous function in $[a,b]$.
For an arbitrary $x_0 \in [a,b]$ you can prove that $\lim_{x \to x_0^+}l(x)=\lim_{x \to x_0^-}l(x)=l(x_0)$ using sequences and the fact that $f$ is a bounded function by definition of a Riemman integrable function.
Also $h(x)=\frac{(f(x)+g(x))-|f(x)-g(x)|}{2}$ which is Riemman integrable as difference of Riemman ntegrable functions on $[a,b]$.
This is because $f(x)-g(x)$ is Riemman integrable thus $|f-g|$ is Riemman integrable.