1. ## Integral Inequalities

Suppose $\displaystyle f$ and $\displaystyle g$ are real valued functions that are Riemann integrable on $\displaystyle [a,b]$. Suppose further that $\displaystyle f(x) \leq g(x)$ for all $\displaystyle x \in [a,b]$ and that a strict inequality holds for at least one point in $\displaystyle [a,b]$. Prove that $\displaystyle \smallint_{a}^{b} f < \smallint_{a}^{b} g$ need not hold. If we add a continuity condition to $\displaystyle f$ and $\displaystyle g$, does the above inequality hold?

Proof. We know that $\displaystyle \smallint_{a}^{b} f \leq \smallint_{a}^{b} g$. Define $\displaystyle f$ and $\displaystyle g$ as follows:
$\displaystyle f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \neq c \\ K \ \ \ \text{if} \ x = c \end{cases}$

$\displaystyle g(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \neq c \\ L \ \ \ \ \text{if} \ x = c \end{cases}$

Here we have $\displaystyle c \in [a,b]$ and $\displaystyle K < L$. Then $\displaystyle \smallint_{a}^{b} f = \smallint_{a}^{b} g = 0$. $\displaystyle \diamond$

If we had a continuity condition, then I think the above inequality will hold? Do we use the following fact: There exists an open interval $\displaystyle (c,d)$ in $\displaystyle [a,b]$ such that $\displaystyle x \in (c,d)$ and $\displaystyle f(u)>0$ for all $\displaystyle u \in (c,d)$? Is this correct?

2. Originally Posted by manjohn12
If we had a continuity condition, then I think the above inequality will hold? Do we use the following fact: There exists an open interval $\displaystyle (c,d)$ in $\displaystyle [a,b]$ such that $\displaystyle x \in (c,d)$ and $\displaystyle f(u)>0$ for all $\displaystyle u \in (c,d)$? Is this correct?
You have the correct idea what contunity. This is similar to the proof of this result: $\displaystyle \int_a^b f = 0 \implies f = 0 \text{ if }f\geq 0 \text{ and continous}$.

Basically what you do is assume that $\displaystyle f(x_0) > 0$ and then you know in an interval of this point we have $\displaystyle f>0$ which leads to a contradiction. You can apply this similar argument to $\displaystyle f-g$.