Prove that $\underline{\int_a^b}f(x)dx=\underline{\int_a^c}f(x)dx+\underline{\int_c^b}f(x)dx.$ Let $f:[a,b]\to \mathbb{R}$ a bounded function. Prove that $$\underline{\int_a^b}f(x)dx=\underline{\int_a^c}f(x)dx+\underline{\int_c^b}f(x)dx,$$ for $a<c<b$.
Let $ \epsilon> 0 $ be given. So, there exists a partition $ \mathcal {P} $ such that $$L (P, f) <\underline {H} (f) + \epsilon,$$ where $ \underline {H} (f) $ is the lower integral. Let's take the following two partitions, $\mathcal {P} _1 = \{a = x_0, x_1, \ldots, x_ {n + 1} = c \}$ and  $\mathcal {P} _2 = \{c = x_ {n + 1}, x_ {n + 2}, \ldots, x_ {2n} = b \}$. Let's call $\mathcal {P} '= \mathcal {P} \cup \{c \}$. Then, $$\underline {H} (a,c)+\underline {H} (c,b)\leq L(\mathcal{P}_1,f)+L(\mathcal{P}_2,f)=L(\mathcal{P}',f)\leq L(\mathcal{P},f).$$
Since $L (P, f) <\underline {H} (f) + \epsilon,$ then $$\underline {H} (a,c)+\underline {H} (c,b)\leq \underline {H} (f) + \epsilon$$
Since $ \epsilon $ is arbitrary, we have one inequality, and the other is almost similar. Is my proof correct? Although I have something weird that is the $ \alpha $ and here it is $ dx $.
 A: Take any partition $P$ of $[a,b]$. If the point $c$ is included, let $P'=P$. Otherwise add the point $c$ to form the refined partition $P'\supset P$.  This induces a partition $P'_1$ of $[a,c]$ and a partition $P'_2$ of $[c,b]$ such that $P' = P'_1 \cup P'_2$,
Hence,
$$L(P,f) \leqslant L(P',f) = L(P'_1,f) + L(P'_2,f) \leqslant  \underline{\int_a}^cf(x) \, dx +\underline{\int_c}^bf(x) \, dx$$
The first inequality holds because $P'$ is  a refinement of $P$. The last inequality holds because the lower integral is the supremum taken over lower sums, e.g.,
$$\underline{\int_a}^cf(x) \, dx = \sup \{L(Q,f): Q \text{ is a partition of  } [a,c] \}$$
Since this holds for every partition $P$ of $[a,b]$, it follows that
$$\underline{\int_a}^bf(x) \, dx = \sup \{L(P,f): P \text{ is a partition of  } [a,b] \} \leqslant \underline{\int_a}^cf(x) \, dx +\underline{\int_c}^bf(x) \, dx $$
Note also that if  $P_1$ and $P_2$ are partitions of $[a,c]$ and $[c,b]$, respectively, then $P = P_1 \cup P_2$ is a partition of $[a,b]$ , and
$$L(P_1,f) + L(P_2,f) = L(P,f) \leqslant \underline{\int_a}^bf(x) \, dx$$
Taking suprema over $P_1$ and $P_2$ on the left it follows that
$$\underline{\int_a}^cf(x) \, dx +\underline{\int_c}^bf(x) \, dx \leqslant \underline{\int_a}^bf(x) \, dx $$
Together both inequalities prove the desired result,
$$\underline{\int_a}^bf(x) \, dx =\underline{\int_a}^cf(x) \, dx +\underline{\int_c}^bf(x) \, dx $$
