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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?

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  • $\begingroup$ If you're asking about why it goes from $<$ to $\leq$, it could just be that they're giving the slightly weaker statement that they need rather than the slightly stronger statement that's possible. If you can see that the first line implies the second line with $<$ replacing $\leq$, then you can see that it also holds for $\leq$, since $\leq$ is weaker than $<$. $\endgroup$ – Kevin Long Dec 11 '17 at 20:21
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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:

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

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

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

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

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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 $L(f)$?

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