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The definition of Riemann-integrability states that if $f$ is Riemann integrable on $[0,1]$ then for any $\epsilon$ there exists a partition $P_\epsilon$ of $[0,1] $ such that for any $P\supset P_\epsilon$ and corresponding choice $T=\{t_i,\dots,t_N\}$, $$\left|\sum_{i=1}^Nf(t_i)(x_i-x_{i-1})-I\right|<\epsilon$$ for some real number $I$.

(We denote the above sum by $S(P,T,f,\alpha)$.)

A sketch of how I think I should prove this would be to fix $\epsilon>0$ so that we have a partition $P_\epsilon=\{x_0=0,x_1,\dots,x_N=1\}$ and then put $\delta<x_1$, and show that for any $c<\delta$, the Riemann sum with the same mesh size as $P_\epsilon$ is very close to $I$, but I am having a hard time formalizing this and maybe that's because I'm not right. Does this seem like a reasonable approach?

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Your idea is good. For given $\varepsilon > 0$ we have to find $\delta > 0$ such that for all $0 < c < \delta$ we have $\left|\int_c^1 f(x) \mathrm{d}x - \int_0^1 f(x) \mathrm{d}x\right| < \varepsilon$. Take the partition $P_\varepsilon$ of $[0,1]$ and $\delta$ as you suggested. For any $0 < c < \delta$, $P_\varepsilon$ gives rise to a partition $P_\varepsilon'$ of $[c,1]$ of smaller or equal mesh size by replacing $x_0 = 0$ by $x_0' = c$. Refine $P_\varepsilon'$ in such a way that for the new partition $P' \supseteq P_\varepsilon'$ we have that all corresponding Riemann sums approximate the integral up to an error of $\varepsilon$. $P = P \cup \lbrace x_{-1} := 0 \rbrace \supseteq P_\varepsilon$ is partition of $[0,1]$. Riemann sums corresponding to $P$ and $P'$ differ by the term $f(t_{-1})(x_0 - x_{i-1})$, whose absolute value is bounded by $\|f\|_\infty \delta$.

An easier way requires a little more knowledge about Riemann integrals, but is considerably shorter: We have $$\left|\int_c^1 f(x) \mathrm{d}x - \int_0^1 f(x) \mathrm{d}x\right| = \left|\int_0^c f(x) \mathrm{d}x\right| \leq \|f\|_\infty c \to 0, \quad c \to 0.$$

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