Question: Prove that for $f$ continuous on $[a,b] \in \Bbb R$, for $c \in [a,b]$, then: $ \int_a^b f(x)\, dx = \int_a^c f(x)\,dx + \int_c^b f(x)\, dx$
My Work:
Let $\{P_k\}$ be a sequence of partitions such that the mesh of $P$ goes to $0$ as $k \to \infty$. For each $k$, denote the points of the partition $P_k$ by $x_0^k < x_1^k < \cdots < x_N^k$. In each subinterval $[x_{i-1}^k, x_i^k]$ choose a point $x_i^{k*}$.
W.L.O.G. let $c \in [a,b]$ be an endpoint of one of the subintervals. If $c$ is not an endpoint, we can refine the partition a point $x_j^k$ at $c$. Then we have: $ \int_a^c f(x)\,dx = \sum_{i=1}^j f(x_i^{k*})(x_i^k - x_{i-1}^k) \quad\text{and}\quad \int_c^b f(x)\, dx = \sum_{i=j+1}^N f(x_i^{k*})(x_i^k - x_{i-1}^k) $ Adding the two partial sums together, we have $\sum_{i=1}^N f(x_i^{k*})(x_i^k - x_{i-1}^k) =\int_a^b f(x)\, dx $
Needing Work: I am not sure I can make the WLOG argument I made, but I'm not sure how else to introduce $c$. And it just seems a little lacking.