Let $$a. The function $$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$$. Why does this equality hold here?
The term $$\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 and $$\pi_R$$ is $$c=x_{n} then $$\pi_L \cup \pi_R$$ is $$a=x_0 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 $$m_k=\inf_{t\in [x_{k-1},x_k]} f(t).$$