1. ## Reimann Integrable

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

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

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 $\int f(x)dx$ (lower integral of f over [0,1]) = $\int f(x)dx$ (upper integral of f over [0,1])

How do I show if it is or not?

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

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

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 $\int f(x)dx$ (lower integral of f over [0,1]) = $\int f(x)dx$ (upper integral of f over [0,1])

How do I show if it is or not?
Notice that on ANY subinterval $[a,b]\subset[0,1]$ that $\sup_{x\in[a,b]}f=3$ and $\inf_{x\in[a,b]}f=1$...so what?

3. so does that mean it's not Riemann integrable because the sup doesn't equal the inf?

4. 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)

5. well.. umm...

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

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

since sup f = 3 and inf f = 1,

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

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

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

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