Absolute value property for solving sums for integrals

Question

I'm working with Darboux integrals for a real analysis class and the following step was presented in the solution for a problem:

For every $\varepsilon > 0$, there exists a partition $\mathcal{P}$ such that:

$\mathopen|L(f;\mathcal{P}) - L(f)\mathclose| < \varepsilon\\L(f)\leq L(f;\mathcal{P}) + \varepsilon$

I'm having a hard time understanding how they go from the absolute value step to the less-than-equal-to step. Is this a property of absolute value? What am I missing in the middle?

Answer 1

Yes, this is a simple property of absolute values. Recall that $|a| < b$ is equivalent to the pair of inequalities $-b < a < b$. Also recall that $a<b$ implies $a \leq b$. Thus we have:

$|L(f;P) - L(f)| < \epsilon$

$-\epsilon < L(f;P) - L(f)$

$L(f) < L(f;P) + \epsilon$

$L(f) \leq L(f;P) + \epsilon$.

Answer 2

Yes, this is just a property of absolute value. You can think about this as coming from the reverse triangle inequality: $|L(f;P)| - |L(f)| \leq |L(f;P) - L(f)| < \epsilon$, which rearranges easily to the result you want.

Alternatively, think about the real number line and notice that if $L(f)$ and $L(f;P)$ are within $\epsilon$ of each other, what is the maximum value of $L(f)$?