Reimann Integrable

**thaopanda** · November 23rd 2009, 04:52 PM

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?

**Drexel28** · November 23rd 2009, 06:37 PM

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?

**thaopanda** · November 24th 2009, 04:05 AM

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

**Drexel28** · November 24th 2009, 07:21 AM

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)

**thaopanda** · November 24th 2009, 09:27 AM

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...

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