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 max and \min_{x\in (a,b)}f(x)=\begin{cases} 0 & \mbox{ if } 1\in (a,b),\\ 1 & \mbox{ otherwise.} \end{cases}
Let n\in \mathbb{N} and let P=(P_i)_{i\in I} be any partition such that P_j=(1-\frac{1}{n},1+\frac{1}{n}) for some j. Clearly U(f;P)=2. Notice that L(f;P)=2-\frac{2}{n}. Since n was chosen arbitrarily, we can get U(f;P)-L(f;P) as small as we want.
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