# Real analysis:Showing a function is integrable

1. Apr 7, 2012

### nlsherrill

1. The problem statement, all variables and given/known data
Let $f:[0,2]-> R$ be defined by$f(x)= 1$ if $x≠1$, and $f(1)= 0$. Show that $f$ is integrable on $[0,2]$ and calculate its integral

2. Relevant equations
Lower integral of f

$L(f)= sup {(P;f): P \in P(I)}$

Upper integral of f

$U(f)= inf {U(P;f): P \in P(I)}$

Where,

$L(P;f)= \sum m_{k}(x_{k}-x_{k-1})$

$U(P;f)= \sum M_{k}(x_{k}-x_{k-1})$

And lastly, $U(f)=L(f)$ if the integral exists

3. The attempt at a solution
So it seems pretty obvious the integral is equal to 2, but I am not sure how to deal with this function at x=1. I tried splitting the interval up into sections, but it turns out partitions only work if the interval is closed and bounded on R, so I couldn't do any open interval stuff. Any ideas?

2. Apr 7, 2012

### Office_Shredder

Staff Emeritus
Given a partition, what can you say about Mk on every sub-interval? What about mk on every sub-interval except for 1?

3. Apr 7, 2012

### nlsherrill

Mk is always 1, and so is mk except for 1 subinterval it will be 0

4. Apr 8, 2012

### HallsofIvy

Staff Emeritus
And the length of every sub-interval goes to 0.

5. Apr 9, 2012

### nlsherrill

The sub-intervals width depends on how large "n" is doesn't it? I don't see what you mean Halls of Ivy.

6. Apr 9, 2012

### Office_Shredder

Staff Emeritus
You're looking at the supremum and infimum over all partitions. Can you sumac partitions where the interval containing 1 had arbitrarily small length?

7. Apr 10, 2012

### nlsherrill

Office,

Are you saying that you could consider the interval containing 1 infintesimally small so that mk and Mk are zero on it? So this would be summing zero for the interval? I am not sure