Real analysis:Showing a function is integrable

[nlsherrill]

Homework Statement

Let f:[0,2]-> R be defined byf(x)= 1 if x≠1, and f(1)= 0. Show that f is integrable on [0,2] and calculate its integral

Homework 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

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?

 

[Office_Shredder]

Given a partition, what can you say about M_k on every sub-interval? What about m_k on every sub-interval except for 1?

 

[nlsherrill]

Given a partition, what can you say about M_k on every sub-interval? What about m_k on every sub-interval except for 1?

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

 

[HallsofIvy]

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

 

[nlsherrill]

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

 

[Office_Shredder]

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

 

[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