According to the definition of integrable, we divide the area into infinite rectangle and each rectangle have same width. Then, we pick up a f(t*) in each subinterval as the height of each rectangle and calculate each rectangle’s area and sum up.

In this case, there are at least 3 subintervals we can make f(t)=0 (at most 6 if t=1, 2, 3 is at the end point of its subinterval).

When the discontinuous point is finite, we can ignore this different of sum because we take the limit as n growing to infinite and there exist another part of sum which is infinite.