Relations between upper and lower Riemann sum

Question

Let , $f$ , $g$ , $h$ be bounded functions on the closed interval $[a,b]$ such that $f(x)\le g(x)\le h(x)$ for all $x\in [a,b]$. Let , $P=\{a=a_0<a_1<a_2<\cdots <a_n=b\}$ be a partition of $[a,b]$. We denote by $U(P,f)$ and $L(P,f)$ the upper sum and lower Riemann sum of $f$ with respect to the partition $P$ and similarly for $g$ and $h$. Which of the following statements is necessarily true ?

(A) If $U(P,h)-U(P,f)<1$ then $U(P,g)-L(P,g)<1$

(B) If $L(P,h)-L(P,f)<1$ then $U(P,g)-L(P,g)<1$

(C) If $U(P,h)-L(P,f)<1$ then $U(P,g)-L(P,g)<1$

(D) If $L(P,h)-U(P,f)<1$ then $U(P,g)-L(P,g)<1$

Since, $f\le g\le h$ so , $U(P,f)\le U(P,g)\le U(P,h)$ and $L(P,f)\le L(P,g)\le L(P,h)$. Combining these two relations we get

$L(P,f)\le U(P,f)\le L(P,g)\le U(P,g)\le L(P,h)\le U(P,h)$.

From this it is clear that ,

$U(P,g)-L(P,g)<U(P,h)-U(P,f)$

$U(P,g)-L(P,g)<L(P,h)-L(P,f)$

$U(P,g)-L(P,g)<U(P,h)-L(P,f)$

$U(P,g)-L(P,g)<L(P,h)-U(P,f)$

So all the options are correct..But it is NOT correct...Where my fallacy ?

Answer 1

The way you combined the inequalities is not correct, as per user137794's counterexample.

Hint: Take $U(P,f) \leq U(P,g) \leq U(P,h)$ as you have it and subtract from each term the number $L(P,f)$. Then use that and the inequality $ L(P,f) \leq L(P,g) \leq U(P,g) $ to deduce an inequality which implies one of the statements in the problem.

Answer 2

It is not true that $U(P,f) \leq L(P,g)$. For example, let $f=x$ and $g=x+1$ on $[0,2]$ and let $P$ simply be $\{0,2\}$. On second thought... let $f=g$.