How does one show that the definite Riemann integral of a function is $ > 0 $? Let $ f \in R[a,b] $, where $ f \ge 0 $, and suppose that $ f(x) > 0 $ for a point of continuity $ x $ of $ f $. Then one can show that
$$
\int_{a}^{b} f(x) ~d{x} > 0.
$$
One can construct an area function $ I(a,b;f) $ with the property
$$
\inf_{x \in [a,b]} f(x) \cdot (b - a) \le I(a,b;f) \le \sup_{x \in [a,b]} f(x) \cdot (b - a).
$$
Since both the infimum and the supremum are greater than $ 0 $, so must be $ I(a,b;f) $.
One can also show that
$$
\int_{a}^{b} |f(x) - g(x)| ~d{x} = 0
$$
if $ f =_{\text{a.e.}} g $.
How does one show this?

Let $x_0$ be the point of continuity. Pick $\epsilon > 0$ so that $f(x_0) > \epsilon > 0$. Since $x_0$ is a point of continuity, one can pick $\delta > 0$ so that:
$$|x - x_0| < \delta \implies |f(x) - f(x_0)| < \frac{\epsilon}{2}$$
But this means that:
$$|x - x_0| < \delta \implies f(x) > f(x_0) - \frac{\epsilon}{2} > \frac{\epsilon}{2}$$
Hence:
$$
\int_a^b f(x) \,dx \ge \int_{x_0-\delta}^{x_0+\delta}f(x) \,dx \ge \frac{\epsilon}{2}((x_0 + \delta) - (x_0 - \delta)) = \epsilon \delta > 0
$$

For the second question, split $[a, b]$ into two sets:
$$A = \left\{ x \in [a, b] : f(x) = g(x)\right\} \\
B = \left\{ x \in [a, b] : f(x) \ne g(x)\right\}$$
Since $f = g$ a.e., $m(B) = 0$ ($B$ is a set of measure $0$). Now, split the integral into two:
$$
\int_{[a, b]} |f(x) - g(x)| dx = \int_A |f(x) - g(x)| dx + \int_B |f(x) - g(x)| dx
$$
Since $f = g$ on $A$, the first integral is $0$. Since $m(B) = 0$, the second integral is $0$. We conclude:
$$
\int_{[a, b]} |f(x) - g(x)| dx = 0
$$
