Doubt in an epsilon delta proof step I am confused with a step in a $\epsilon$-$\delta$ proof. Let's consider $[a,b]$. Let's consider a partition of this interval the usual way with Riemann integrals. Let $f:[a,b]\to\mathbb{R}$ be a continuous function. Consider an that the following inequality holds
$$
\bigg|\int_a^b\,|f(x)|\,dx-\sum_{k=1}^n|f(x)|(t_k-t_{k-1})\bigg|<\epsilon
$$
using the triangle inequality $|a|-|b|\leq|a-b|$ we can write
$$
\int_a^b\,|f(x)|\,dx-\sum_{k=1}^n|f(x)|(t_k-t_{k-1})\leq\bigg|\int_a^b\,|f(x)|\,dx-\sum_{k=1}^n|f(x)|(t_k-t_{k-1})\bigg|<\epsilon
$$
this implies
$$
\int_a^b\,|f(x)|\,dx<\sum_{k=1}^n|f(x)|(t_k-t_{k-1})+\epsilon
$$
nonetheless my my book writes $\leq$ instead of $<$
$$
\int_a^b\,|f(x)|\,dx\leq\sum_{k=1}^n|f(x)|(t_k-t_{k-1})+\epsilon
$$
why not the strict inequality?
 A: Due to the arbitrary choice of $\epsilon$, the last two inequalities in the question body are in fact equivalent.
To simplify writing, restate the triangle inequality in the question body as
$$|y|-|z|\leq|y-z|,$$
where
\begin{align}
y &= \int_a^b\,|f(x)|\,dx \\
z &= \sum_{k=1}^n|f(x)|(t_k-t_{k-1})
\end{align}
The given condition is that for any $\epsilon > 0, |y-z| < \epsilon$, and OP asks for difference between
$$y < z + \epsilon \quad \forall\, \epsilon > 0 \tag1 \label1$$
and
$$y \le z + \epsilon \quad \forall\, \epsilon > 0. \tag2 \label2$$
They are in fact equivalence: \eqref{1} clearly implies \eqref{2}.  To show the converse, fix any $\epsilon > 0$, and choose a smaller $\epsilon' = \epsilon/2$ in \eqref{2}, so that $y \le z + \epsilon' = z + \epsilon/2 < z + \epsilon$.  Since the choice of $\epsilon$ is arbitrary, \eqref{1} is satisfied.

For the author's choice of $\le$ instead of $<$, IMHO, it is because proving $c \le d$ is often easier than proving $c < d$: in the later, some care is sometimes needed to exclude the case $c = d$.  However, in this question, since the choice of $\epsilon$ is arbitrary, such difference doesn't exist.
