Monotonocity of Integrals, if and only if? Assume that $f(x)$ and $g(x)$ are Riemann Integrable and that $\int_a^W f(x) \, dx<\int_a^W g(x) \, dx$ for all $W \in [a,b]$.
Does it follow that
$$ f(x) < g(x)  $$
for all $W \in [a,b]$.
Is it true with weak inequalities?
 A: I think this doesn't hold in general. Fix some $c \in [a,b]$ and take for example $g : [a,b] \to \mathbb{R}$, with $g(x)=1$, and
$$
f : [a,b] \to \mathbb{R}, \quad x \mapsto 
\begin{cases}
0 & x \in [a,b]\setminus \{c\}, \\
2 & x=c.
\end{cases}
$$
Then $f$ and $g$ are Riemann integrable on $[a,b]$, because $g$ is continuous and $f$ is bounded and has only one discontinuity. We indeed have that
$$
\int_a^W f(x) \, dx < \int_a^W g(x)\, dx
$$
for all $W \in [a,b]$. However, for $x=c$, we have $f(c)=2 > 1 = g(c)$.
A: No, because you can modify the value of the function at a single point (and hence finitely many points) for example without changing the value of the integral.
In fact, the converse fails quite badly. To demonstrate this, note that by defining $h=g-f$, we can rephrase your question equivalently as

Suppose $h:[a,b]\to\Bbb{R}$ is Riemann-integrable and for every $w\in (a,b]$, we have $0<\int_a^wh(x)\,dx$. Does it follow that $h\geq 0$ on $[a,b]$?

What we can do is that for any $\epsilon>0$, we can find a continuous function $h:[a,b]\to\Bbb{R}$ such that for every $w\in (a,b]$, we have $0<\int_a^wh(x)\,dx$, but such that $h<0$ on $(a+\epsilon,b]$. In words, we can ensure that $h$ has positive integral everywhere, even though it is mostly negative.
Why is this possible? The idea is very simple, given $\epsilon>0$, think of $h$ as having a very tall spike between $a$ and $a+\epsilon$, and then after $a+\epsilon$, make $h$ be a constant negative value, which is very very small in absolute value. See the picture below; I leave it to you to transcribe the picture into exact formulas if you wish.


On the other hand, you can ask the following question:

Suppose $h:[a,b]\to\Bbb{R}$ is Riemann-integrable and for any $c,d\in [a,b]$ we have $\int_c^dh(x)\,dx \geq 0$. Does it follow $h\geq 0$ on $[a,b]$?

The answer is "almost". The set of points where $h<0$ must now be very small, or more precisely, it must have Lebesgue measure zero. (Note that because we are now allowing both the upper and lower limits to vary, the picture above no longer applies because if in the picture above we take $c,d$ ot be larger than $a+\epsilon$ then the integral is negative).
A slightly simpler thing to prove is this:

If $h:[a,b]\to\Bbb{R}$ is continuous and for every $c,d\in[a,b]$ we have $\int_c^dh(x)\,dx\geq 0$, then $h\geq 0$ on $[a,b]$.

The contrapositive of this statement is easy to prove: if $h$ is strictly negative at some point, then by continuity, there is an open neighborhood of that point on which $h$ is strictly negative, so integrating over this region yields a strictly negative number.
