Let
$$ f(x) = \begin{cases}
1, & 0 \le x < 1 \\
0, & x=1 \\
1, & 1
\end{cases}$$
How Prove that f(x) integrable on [0,2]
I know that f(x) defined and Bounded on [a,b] then f(x) integrable
iff
∀ϵ>0 there is Partition P so that
U(f;P)−L(f;P)<ϵ
how can i prove that?
thanks
Answer
Notice that on any interval (a,b)⊂[0,2] we have that maxx∈(a,b)f(x)=1 and minx∈(a,b)f(x)={0 if 1∈(a,b),1 otherwise.
Let n∈N and let P=(Pi)i∈I be any partition such that Pj=(1−1n,1+1n) for some j. Clearly U(f;P)=2. Notice that L(f;P)=2−2n. Since n was chosen arbitrarily, we can get U(f;P)−L(f;P) as small as we want.
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