2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $m_k=\inf_{t\in [x_{k-1},x_k]} f(t).$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.