Step in proof on Riemann Sums from Spivak Calculus.

Question

I was working out a proof in Spivak's Calculus (2008) - pg 279. The following is a screenshot of the portion of the proof I'm having trouble with.

My question is in working out combining steps 1,2, and 3 correctly. I want to arrive at

$$\bigg|\sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x)dx \bigg| < \epsilon \\ \Rightarrow\ -\epsilon < \sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x)dx < \epsilon$$

Fiddling around with equation 2, I would get something of the form

$$ 0 \leq \sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - L(f,P) \leq U(f,P) - L(f,P) < \epsilon$$

The same would occur for $\int_{a}^{b}f(x) dx$. Now using this idea I get something of the form:

$$\epsilon > U(f,P) - L(f,P) \geq \sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - L(f,P) \geq \sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x) \geq ?? $$

Here is my issue, I can't say for sure that $\sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x) \geq 0$. Nothing that I have can imply such and as a result of this I can't conclude that $\sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x) > -\epsilon$. Which would allow me to finish this part of the proof. From experience I know it is a minor algebraic thing I'm missing, but I suppose I'm mentally fatigued and not seeing it. Some assistance would be nice.

Answer 1

Hint: Multiply equation $(3)$ by $-1$ and add to equation $(2)$ to get:
$-(U(f,p) -L(f,P))\leq -\int_{a}^{b}f(x)dx+\sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) \leq U(f,P) - L(f,P) $
In other words, we have $-\epsilon\lt y\lt \epsilon$, whence $|y|\lt \epsilon$

Answer 2

$(2)$ and $(3)$ imply that both the sum and the integral are between $L(f,P)$ and $U(f,P)$ so the absolute difference between them cannot be more than $U(f,P)-L(f,P)$ and by $(1)$ this latter expression is less than $\epsilon.$