# Reimann Integrable

• Nov 23rd 2009, 04:52 PM
thaopanda
Reimann Integrable
Determine whether the function f: [0,1] $\displaystyle \rightarrow$ R given by:

f(x):=
3 if $\displaystyle x \in Q \cap [0,1]$
1 if $\displaystyle x \in (R$ \ $\displaystyle Q) \cap [0,1]$

is Riemann integrable on [0,1].

I know that if it is Riemann integrable, sup L(P, f) = inf U(P,f), P being the partition of [0,1]
or $\displaystyle \int f(x)dx$ (lower integral of f over [0,1]) = $\displaystyle \int f(x)dx$ (upper integral of f over [0,1])

How do I show if it is or not?
• Nov 23rd 2009, 06:37 PM
Drexel28
Quote:

Originally Posted by thaopanda
Determine whether the function f: [0,1] $\displaystyle \rightarrow$ R given by:

f(x):=
3 if $\displaystyle x \in Q \cap [0,1]$
1 if $\displaystyle x \in (R$ \ $\displaystyle Q) \cap [0,1]$

is Riemann integrable on [0,1].

I know that if it is Riemann integrable, sup L(P, f) = inf U(P,f), P being the partition of [0,1]
or $\displaystyle \int f(x)dx$ (lower integral of f over [0,1]) = $\displaystyle \int f(x)dx$ (upper integral of f over [0,1])

How do I show if it is or not?

Notice that on ANY subinterval $\displaystyle [a,b]\subset[0,1]$ that $\displaystyle \sup_{x\in[a,b]}f=3$ and $\displaystyle \inf_{x\in[a,b]}f=1$...so what?
• Nov 24th 2009, 04:05 AM
thaopanda
so does that mean it's not Riemann integrable because the sup doesn't equal the inf?
• Nov 24th 2009, 07:21 AM
Drexel28
Quote:

Originally Posted by thaopanda
so does that mean it's not Riemann integrable because the sup doesn't equal the inf?

You tell me. And justify your answer. (I will verify it if you do)
• Nov 24th 2009, 09:27 AM
thaopanda
well.. umm...

so L(P,f) = $\displaystyle \sum m_{k}(x_{k}-x_{k-1})$ where $\displaystyle m_{k}$ = inf f(x)

and U(P,f) = $\displaystyle \sum M_{k}(x_{k}-x_{k-1})$ where $\displaystyle M_{k}$ = sup f(x)

since sup f = 3 and inf f = 1,

L(P,f) = $\displaystyle \sum (x_{k}-x_{k-1})$
U(P,f) = $\displaystyle \sum 3(x_{k}-x_{k-1})$

and $\displaystyle (x_{k}-x_{k-1}) \leq 1$ because that's the maximum partition in [0,1]

so,
sup L(P,f) $\displaystyle \leq$ 1
inf U(P,f) $\displaystyle \leq$ 3

but the sup f = 3... yeah, I'm lost... (Crying)