Claim: A function with removable discontinuities is integrable.

• What does having a removable discontinuity mean?

A function f(t) has a removable discontinuity at a if lim t→a and f(a) are defined but not equal.

• What does integrable mean?

A function f(t)  is integrable on an interval [a,b] if the lim n→∞ of the sum of the rectangles (height= f(t*) × width=(a-b)/n) is the same no matter the choice of t*. 

t* is a point taken from one interval.

Given the fact that all continuous functions are integrable a function like f(t)=3  is integrable and for any rectangle the height=3.

Lets  calculate the integral of f(t)=3 from [0,5]. We divide the interval into equal length subintervals and multiply them by a sample point t* of height=3. The area under the curve=6. We can also get to this result by analyzing the figure that x=0, x=5, y=3, and the t-axis make. A rectangle and we know the area of a rectangle (A=width×height)

Now lets calculate the area under the curve for the same interval [0,5] for a similar function with a removable discontinuity

f(t)=    3 when t≠3   and   =5 when t=3

Again we divide the interval into equal length subintervals and multiply them by a sample point t*. Now not all t* will give f(t*)=3. We can pick t* to give f(t*)=5 and the area of that rectangle will be different from the rest of the rectangles. However, consider the difference that this different rectangle will have on the total sum, is it significant? In the case of our rectangle, does the rectangle stop being a rectangle after taking a single point out? No, it doesn’t. The discontinuity is negligible. And what if the function/rectangle has two, three, four, or ten points of discontinuity? It is still a rectangle. But what if the function has an infinity number of discontinuity points? Then, we can make all of our t* to give a different value and the area will not be the same.

Provided that the number of discontinuity points is finite, f(t) is integrable on the interval [a,b]