# Integrator as an average?

1. Mar 6, 2014

### sammycaps

Hello all,

I'm going through Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger, and I read a curious line that I was hoping someone here could clear me up on (perhaps I'm thinking too much into it).

The begininning of the section on the Lebesgue integral introduces a sequence of Riemann integrable functions which converge (somehow) to a function, ƒ,which is not Riemann integrable. The authors go on to write....

"Why is ƒ not Riemann integrable? The fault lies not with the average used, namely α(x)=x, but rather with the sets averaged over, i.e., intervals."

where α(x) is the integrator. Generally, when I think of α(x), I think of it as a weighting function. So, what do the authors here mean that α(x)=x is the "average used"?

2. Mar 7, 2014

### Stephen Tashi

I don't have copy of that book. You should state the example.

One guess is that the quotation is an informal reference to the fact a net quantity (like total dollars) can be computed from the total number of things involved ( like total people) times the average of that quantity per thing (like "mean cost per person"). An average can be viewed as a constant "weight" that makes a product yield the desired result.