A case in which the lower Riemann integral is additive Let $A$ and $B$ be two open subsets of $[a,b]$.
I claim that
$$\underline {\int_a^b} \chi_A + \chi_B = \underline {\int_a^b} \chi_A + \underline {\int_a^b} \chi_B$$ where $\chi_A$ is the characteristic function of $A$.
Since the lower Riemann integral is superadditive, I should prove that $$\underline {\int_a^b} \chi_A + \chi_B \le \underline {\int_a^b} \chi_A + \underline {\int_a^b} \chi_B$$
Do I have to use the Heine-Borel theorem in some way ?
Many thanks in advance.
 A: Suppose $\underline{\int_{[a, b]}}(\chi_A+\chi_B)>\underline{\int_{[a, b]}}\chi_A+\underline{\int_{[a, b]}}\chi_B$, choose a partition $\mathcal{P}_1: a=t_0<t_1<\dots<t_n=b$ whose corresponding lower sum $L_1$(of $\chi_A+\chi_B$) lies between them.
Call an interval $[r, s]$ problematic if
$$\min_{[r, s]}(\chi_A+\chi_B)>\min_{[r, s]}\chi_A+\min_{[r, s]}\chi_B.$$
Observe that if $[r, s]$ is problematic, then every point of $[r, s]$ must be in one of $A, B$ and at the same time it contains points from $A\setminus B, B\setminus A$. In particular, the left side above should be $1$.
In the partition above, some $[t_i, t_{i+1}]$ should be problematic, else the lower sum of $\chi_A+\chi_B$ will be equal to the sum of lower sums of $\chi_A, \chi_B$. Let $P_1$ be the collection of problematic intervals in $\mathcal{P}_1$.
Subdivide elements of $P_1$ in half to arrive at a new partition $\mathcal{P}_2$ with lower sum $L_2$ of $\chi_A+\chi_B$. Since $L_1\leq L_2, \mathcal{P}_2$ must also contain some problematic intervals, and every problematic interval of $\mathcal{P}_2$ must be contained in a problematic interval of $\mathcal{P}_1$. Let $P_2$ be the collection of problematic intervals in $\mathcal{P}_2$.
Inductively define $\mathcal{P}_n, L_n, P_n, n\geq 1$. The collection $\cup_{n\geq 1}P_n$ is partially ordered by inclusion and contains chains of any finite length (simply take an element of $P_n$ and go backwards to obtain a chain of length $n$) and finitely many chains of a given length (because each $P_n$ is finite).
Suppose $C_1\supset C_2\supset\dots$ is an infinite chain in $\cup_{n\geq 1}P_n$. Note that each $C_i$ is a problematic interval and since we are halving the lengths, the lengths of these intervals decreases to $0$. Therefore, $\cap_{n\geq 1} C_n$ is a singleton $\{z\}$ (we need an infinite chain for this to work). Since each $C_i$ is problematic, $z\in A\cup B$ and every $C_i$ has a point from $A\setminus B$ and one from $B\setminus A$.
Thus, we have a sequence in $A\setminus B$ converging to $z$ and a sequence in $B\setminus A$ also converging to $z$. Since $z\in A\cup B$, we have a contradiction as $A, B$ are open.
Now, why does an infinite chain exist? See Given an infinite poset, show that it contains either a infinite chain or an infinite totally unordered set. or Prove that for any infinite poset there is an infinite subset which is either linearly ordered or antichain.
Therefore the lower integrals must be equal.
