# In what situations the sum of darboux sums can be

Say you have two functions $f,g:[a,b] \rightarrow \mathbb{R}$, both integrable. Let $P$ be a partition of $[a,b]$, $P = (x_0,\ldots,x_n)$. Let's denote the lower Darboux integral of a function $f$ with $L_ {f,P} = \sum\limits_{i=1}^{n}(x_i-x_{i-1})m_i,$ where $m_i =\inf\limits_{x\in[x_{i-1},x_i]} f(x)$.

My Math professor wrote that $L_{f,P} + L_{g,P} \leq L_{f+g,P}$. I can see how these two are equal, but I can't find an example where the left term is smaller than the right one. Can you think of one? Maybe he's wrong?

• Consider $f(x) = x$ and $g(x) = -x$. – Daniel Fischer Jul 1 '17 at 14:33

Take the functions $f ,g$ defined at $[x_j,x_{j+1}]$ by
$$f (x_j)=0$$ $$x_j <x\le x_{j+1}\implies f (x)=2$$ $$g (x_{j+1})=0$$ $$x_j\le x <x_{j+1} \implies g (x)=2$$
$$\min f=\min g=0$$
$$\min (f+g)=2$$
This comes from $$\inf \{f (x)+f (y)\}\ne \inf \{f (x)+g(x)\}$$
We can take for example: $f,g:[0,2]\rightarrow\mathbb{R},P=\{0,1,2\}$, $f(x) = \begin{cases} 0 & \text{x}\neq1 \\ -1 & \text{x=1} \end{cases}$ $g(x) = \begin{cases} 0 & \text{x}\neq1 \\ 1 & \text{x=1} \end{cases}$ then: $$L_{f,p}=-2,L_{g,p}=0,L_{f+g,p}=0$$ and exists: $$L_{f,p}+L_{g,p}=-2<0=L_{f+g,p}$$