# Riemann-Darboux Integrability of Subinterval

I'm studying Riemann-Darboux integration. I'm trying to prove the following rather intuitive notion for integrals. Please let me know if you find any errors in this proof, as I'm self-studying this topic.

Theorem: Suppose $$f$$ is Riemann-Darboux integrable on $$[a,b]$$. Let $$c\in(a,b)$$. Then, $$f$$ is Riemann-Darboux integrable on the intervals $$[a,c]$$ and $$[c,b]$$.

Attempted Proof: Since $$f$$ is Riemann-Darboux integrable on $$[a,b]$$, for arbitrary $$\epsilon>0$$, there exists a partition $$P$$ of $$[a,b]$$ such that $$U(f,P)-L(f,P)<\epsilon$$.

Let $$n_p,n_{p*}$$, and $$n_{p'}$$ be the number of partition parts in $$P$$, $$P^*$$, and $$P'$$, respectively. Also, let $$m_i=\inf_{x\in[x_{i-1},x_i]}f(x)$$ and $$M_i=\sup_{x\in[x_{i-1},x_i]}f(x)$$.

Consider the partition of $$[a,c]$$ given by $$P^*=P\cap[a,c]$$. Then, $$U(f,P^*)-L(f,P^*)=\sum_{i=1}^{n_{p*}}(M_i-m_i)\Delta x_i\le\sum_{i=1}^{n_{p*}}(M_i-m_i)\Delta x_i+ \sum_{n_{p*}+1}^{n_p}(M_i-m_i)\Delta x_i=\sum_{i=1}^{n_p}(M_i-m_i)\Delta x_i=U(f,P)-L(f,P)<\epsilon$$ Therefore, $$f$$ is integrable on $$[a,c]$$.

Next, consider the partition $$[a,b]$$ given by $$P'=P\cap[c,b]$$. Then,

$$U(f,P')-L(f,P')=\sum_{i=n_{p*}+1}^{n_{p'}}(M_i-m_i)\Delta x_i\le\sum_{i=1}^{n_{p*}}(M_i-m_i)\Delta x_i+ \sum_{n_{p*}+1}^{n_p}(M_i-m_i)\Delta x_i=\sum_{i=1}^{n_p}(M_i-m_i)\Delta x_i=U(f,P)-L(f,P)<\epsilon$$ Therefore, $$f$$ is integrable on $$[c,b]$$. $$\square$$

Any and all feedback, or alternative proofs are appreciated. I love to see different arguments to expand my skill set.

This is essentially fine, although you are tacitly assuming that $$c$$ itself is a member of the partition, $$P$$. This isn't a big deal though - a $$P$$ such that $$U(P,f)-L(P,f)<\epsilon$$ is guaranteed by integrability, and you can always just add $$c$$ to this partition if it is not already there. Namely, if $$P_{c}$$ is this new partition, called a refinement of $$P$$, one must have $$U(P_{c},f)-L(P_{c},f)\leq U(P,f)-L(P,f)<\epsilon$$ and you can proceed as you have done in your proof.
An alternative approach is given by Lebesgue's Criterion for Riemann Integrability, which states that Riemann/Darboux Integrability is equivalent to boundedness plus continuity up to a null set (see Wikipedia for a good explanation of null sets). Since your functions is integrable on $$[a,b]$$, it is bounded and continuous up to a null set on $$[a,b]$$, and hence, on $$[a,c]$$ and $$[c,b]$$ as well.