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Let $a<c<b$. The function $f$ is integrable over $]a,b[ \iff f$ is integrable over $]a,c[$ and $]c,b[$. We get: $\int_a^bf=\int_a^cf+\int_c^bf $

In the proof of this theorem ($\Leftarrow$), we take an arbitrary partition $\pi_L$ of $]a,c[$ and a partition $\pi_R$ of $]c,b[$. Then we consider the partition $\pi := \pi_L \cup \pi_R$ of $]a,b[$.

We get that $s_{\pi_L}+s_{\pi_R} = s_\pi \le \underline{\int_a^b} f$. Why does this equality hold here?

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The term $\underline{\int_a^b} f(x)\,dx$ is defined as the supremum of the set $s_{\pi}$ where $\pi$ is a generic partition of the interval $[a,b]$. In particular $\pi_L \cup \pi_R$ is a partition of $[a,b]$ and $$s_{\pi_L \cup \pi_R}\leq \underline{\int_a^b} f(x)\,dx.$$ Moreover, if $\pi_L$ is $a=x_0<x_1<\dots<x_n=c$ and $\pi_R$ is $c=x_{n}<x_{n+1}<\dots<x_{n+m}=b$ then $\pi_L \cup \pi_R$ is $a=x_0<x_1<\dots<x_{n+m}=b$ and $$s_{\pi_L}+s_{\pi_R}=\sum_{k=1}^n m_k(x_k-x_{k-1})+\sum_{k=n+1}^{n+m} m_k(x_k-x_{k-1})=\sum_{k=1}^{n+m}m_k(x_k-x_{k-1}) =s_{\pi_L \cup \pi_R}$$ where $m_k=\inf_{t\in [x_{k-1},x_k]} f(t).$

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  • $\begingroup$ @Zachary Any further doubts? $\endgroup$
    – Robert Z
    Jan 12, 2019 at 8:40
  • $\begingroup$ No, definitely not! Very clear explanation. Thank you! $\endgroup$
    – MyWorld
    Jan 12, 2019 at 8:44

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