Is a Function with One Discontinuity Still Integrable on a Closed Interval?

[ptolema]

Homework Statement

suppose that f is bounded on [a,b] and that f is continuous at each point in [a,b] with the exception of x₀ in (a,b). prove that f is integrable on [a,b].

Homework Equations

f is continuous on [a,x₀] and [x₀,b]

The Attempt at a Solution

intuitively, i know that this is true because i can envision a hole or jump discontinuity and still find the area under f. still, the theoretical approach is daunting. how do i approximate the area for the regions [a,x₀] and [x₀,b]. is it enough to say that for some subinterval containing x₀, the area is approx. 0?

 

[Tedjn]

Is it enough to say that for some subinterval containing x0, the area is approx. 0?

It's probably not enough to say this if you are to be truly rigorous. However, this is the intuition. As subintervals surrounding x₀ become smaller and smaller, the area of some rectangle gets smaller. But what area exactly? And why is it important that f is bounded?

To start answering these questions rigorously, you need to go back to your definition of the integral (I assume the Riemann integral?). At first, it may seem daunting, but if you think about it for awhile, I hope that you'll become so familiar with the ideas/definitions that you won't feel overwhelmed.

 

[JonF]

If you use the oscillation definition of continuity it might make this problem easier

 

[JonF]

or you could show ∫_a^bf(x)dx = ∫_a^x0f(x)dx + ∫_x0^bf(x)dx fairly easily too I suppose by breaking it into two cases of the type of discontinuity possible in a bound set. The other method is better though, since it could much more easily be generalized to sets with multiple discontinuities.