Prove that if $g(x):=0$ for $0\le x\le1/2$and $g(x):=1$ for $1/2\lt x\le1$ then the Darboux Integral of $g$ on $[0,1]$ is equal to $1/2$.
My answer: Lets define a partition $P_\epsilon =(0,1/2,1/2+\epsilon,1)$ thne the upper sum $U(g,P_\epsilon)=0(1/2-0)+1(1/2+\epsilon-1/2)+1(1-(1/2+\epsilon)=0+\epsilon+1/2-\epsilon=1/2$ There fore the upper integral satisfies $U(g)\le1/2$. Similarly the lower sum $L(g,P_\epsilon)=0+0+1/2-\epsilon=1/2-\epsilon$. There fore the lower integral satisfies $L(g)\ge1/2$. We now have $1/2\le L(g)\le U(g)\le1/2$ and $L(g)=U(g)$. thus the Darboux integral of g is $\int_0^1 g=1/2$
Does the conclusion hold if we change the value of g at the point 1/2 to 13? I'm not sure what exactly it is asking me to do. If I change the 1/2 in the intervals I get $U(g)\le-12$ and $L(g)\ge-12-\epsilon$ which does not hold. Is that correct?