Dissection and Integrals

Question

Let $[a,b]$ be an interval. Let $D_1,D_2$ be arbitrary dissections of $[a,b]$. Then $U(f,D_1 \cup D_2)$ $=$ $U(f,D_1)+U(f,D_2)$.

My proof so far: Suppose $D_1$ $=$ $\{$ $a,x_1,x_2.....,x_n=b$ $\}$ and $D_1$ $=$ $\{$ a,$a_1,...,a_m=b$ $\}$ . Clearly, $U(f,D_1)+U(f,D_2) \leq U(f,D_1\cup D_2)$. If $m=n$ then $U(f,D_1)$ $+$ $U(f,D_2)$ $=$

$\sum^{n}_{i=1}$ $(x_i-x_{i-1})sup_{[x_{i-1},x_i]}f$ $+$ $\sum^{n}_{i=1}$ $(a_i-a_{i-1})sup_{[a_{i-1},a_i]}f$

Then I must check the case for $m<n$ How do I complete this proof? Further

Answer 1

The statement is not true.

Take the interval $[0,1]$ and $D_1=\{0,\frac{1}{2},1\}$ and $D_2=\{0,\frac{1}{4},1\}$ and $f(x)=x$

$D_1 \cup D_2=\{0,\frac{1}{4},\frac{1}{2},1\}$

$U(f,D_1 \cup D_2)=\frac{1}{4}\frac{1}{4}+\frac{1}{2}\frac{1}{4}+\frac{1}{2}<1$

$U(f,D_1)+U(f,D_2)=(\frac{1}{2}\frac{1}{2}+\frac{1}{2})+\frac{1}{4}\frac{1}{4}+\frac{3}{4}>1$

If $D_1 \cap D_2=\emptyset$ then i believe it is true.