If $f(x) < g(x)$, prove that $\int_a^b f(x) dx < \int_a^b g(x)dx.$ 1) Let $f$ and $g$ be Riemann integrable functions on $[a,b]$. Suppose that $f(x) < g(x)$ for each $x\in [a,b]$. Prove that $\int_a^b f(x) dx < \int_a^b g(x)dx.$
Basically my idea was to break the integral down into partitions and show the inequality for each... here's my work... not sure if it's correct
Let $f$ and $g$ be Riemann integrable functions on $[a,b]$ and suppose that $f(x) < g(x)$ for each $x\in [a,b]$. Since $f$ and $g$ are Riemann integrable, we know that the points at which they are discontinuous form a zero set. Then $\exists A \subseteq [a,b]$ such that A is not a zero set and $\forall z\in A$, $f$ and $g$ are continuous at $z$. Suppose $x_0\in A$. Then $f(x_0) < g(x_0)$, since $x_0$ is in the domain of $f$ and $g$. Now let $\epsilon = g(x_0) - f(x_0)$. By the continuity of $f$ and $g$ at $x_0$, we have that $\exists \delta > 0 $ such that if $x\in (x_0 - \delta, x_0 + \delta)$ then $\mid f(x_0) - f(x) \mid < {\epsilon \over 4.}$ Then $f(x_0) - {\epsilon \over 4} < f(x) < f(x_0) + {\epsilon \over 4}$. Similarly, $g(x_0) - {\epsilon \over 4} < g(x) < g(x_0) + {\epsilon \over 4}$. Then $(g(x_0) - {\epsilon \over 4}) - (f(x_0) + {\epsilon \over 4}) < g(x) - f(x)$, so $f(x) + {\epsilon \over 2} < g(x) $. Now take $$\int_a^b g(x)dx = \int_a^{x_0 - \delta} g(x)dx + \int_{x_0 - \delta}^{x_0 + \delta} g(x)dx + \int_{x_0 + \delta}^b g(x)dx. $$ We know $$\int_a^{x_0 - \delta} f(x)dx \leq \int_a^{x_0 - \delta} g(x)dx,$$  $$ \int_{x_0 + \delta}^b f(x) dx \leq  \int_{x_0 + \delta}^b g(x) dx, $$ and $$ \int_{x_0 - \delta}^{x_0 + \delta} (f(x) + {\epsilon \over 2})dx \leq  \int_{x_0 - \delta}^{x_0 + \delta} g(x)dx. $$ But \begin{align} \int_{x_0 - \delta}^{x_0 + \delta} (f(x) + {\epsilon \over 2})dx & = \int_{x_0 - \delta}^{x_0 + \delta} f(x) dx + \int_{x_0 - \delta}^{x_0 + \delta}  {\epsilon \over 2} dx\\
&=  \int_{x_0 - \delta}^{x_0 + \delta} f(x) dx + {\epsilon \over 2}\mid 2\delta \mid.  \end{align} Thus $$ \int_{x_0 - \delta}^{x_0 + \delta} f(x) dx < \int_{x_0 - \delta}^{x_0 + \delta} g(x) dx.$$ Thus $\int_a^b f(x) dx < \int_a^b g(x)dx.$
Is this ok?
 A: Hint: $h:=g-f>0$ on $[a,b]$. Can you say that
$$\int_a^b h(x)dx>0$$
by definition of the Riemann integral?
A: Let $h(x) = g(x) - f(x)$ then $h(x)$ is Riemann-integrable on $[a, b]$ and hence is continuous almost everywhere. Thus there must be an interior point $c \in (a, b)$ where $h(x)$ is continuous. Since $h(c) > 0$ it follows that there is an interval $[c -\delta, c + \delta]$ where $h(x)$ is positive and thus $m_{c} = \inf \{h(x)\mid x \in [c -\delta, c + \delta]\} > 0$. Clearly we have $$\int_{a}^{b}h(x)\,dx \geq \int_{c - \delta}^{c + \delta}h(x)\,dx \geq 2\delta\cdot m_{c} > 0$$ and hence $$\int_{a}^{b}f(x)\,dx < \int_{a}^{b}g(x)\,dx$$
Update: Looking at OP's comment to my answer even I figured out a much simpler solution which avoids Lebesgue's theorem ("a bounded function is Riemann integrable if and only if it is continuous almost everywhere"). This may or may not be what OP had in his mind.
Since $h(x)$ is integrable over $[a, b]$ both lower and upper sums for $h(x)$ converge to same value $A$. And clearly since $h(x) > 0$ we must have lower sums as non-negative so that $A \geq 0$. Suppose $A = 0$. Then the upper sums also converge to $0$. It follows that the upper bound of $h(x)$ must be $0$ in some sub-interval. This contradicts that $h(x) > 0$. Hence we must have $A > 0$ i.e. $\int_{a}^{b}h(x)\,dx > 0$.

Update: The argument in the preceding paragraph is wrong. The proper proof is not that easy. Do have a look at this answer.
