If f is a function defined on [a,b] which is not continous,then L(P,f)> U(P,f) for any partition P of [a,b].State true or false. Give reason?
false..
Theorem: Let $\displaystyle f: [a,b] \rightarrow \mathbb{R}$ be a bounded function. Then $\displaystyle L(f)$ and $\displaystyle U(f) $ exist; and $\displaystyle L(P,f) \leq L(f) \leq U(f) \leq U(P,f)$..
the theorem does not have any condition on $\displaystyle f$ to be continuous or not, thus it holds for any bounded function $\displaystyle f: [a,b] \rightarrow \mathbb{R}$
if you have not seen that theorem, you may use a counter-example (Modified Dirichlet Function).
$\displaystyle f(x) = 1$ if $\displaystyle x\in \mathbb{Q} \cap [0,1]$
$\displaystyle f(x) = 0$ if $\displaystyle x\in \mathbb{Q}^\prime \cap [0,1]$
and show that L(P,f) < U(P,f)