14.12. THE VITALI COVERING THEOREMS 341

Proof: Since f is continuous, there is a sequence of step functions {sn} which con-verges uniformly to f . In fact, you could partition [a,b] as a = xn

0 < · · · < xnn = b and

consider

sn (x) =mn

∑k=1

f (xk−1)X[xk−1,xk) (x) , |xk− xk−1|=b−a

n

Then if you consider the generalized Riemann integral of sn (x) it equals∫ b

asn (x)dF =

n

∑k=1

f (xk−1)(F (xk−)−F (xk−1−)) =∫

sndµF

The integrals on the left converge as n→∞ to∫ b

a f dF where this is the generalized Riemannintegral of f because of the uniform convergence of the step functions to f . But this equalsthe Riemann Stieltjes integral since f is continuous. On the right, you get convergence to∫

f dµF again because of uniform convergence.This shows that the new integral coincides with the generalized Riemann Stieltjes in-

tegral on all continuous functions continuous on a closed interval which is the same as theRiemann Stieltjes integral. They give the same answer on continuous functions. The ad-vantage of the Lebesgue integral is that it has all those wonderful convergence theoremswhich are particularly easy to understand and use. You don’t have these theorems withthe Riemann Stieltjes integral and the version of these convergence theorems which holdfor the generalized Riemann integral are much more difficult to prove. Compare the easymonotone convergence theorem with the much more difficult one in the Chapter on thegeneralized Riemann integral.

Also, there is the following definition of the integral over a measurable subset.

Definition 14.11.8 Let (Ω,F ,µ) be a measure space and let E be a measurablesubset of Ω. Then if | f XE | is integrable,

∫E f dµ ≡

∫XE f dµ .

When we are considering the Lebesgue Stieltjes measures, we might write the followingfor f Borel measurable: ∫ b

af dµF ≡

∫X[a,b] f dµF

Of course the notation on the right is preferable because you might have µF (a) > 0. Thenotation on the left might seem a little ambiguous about whether the end points are consid-ered. Of course the main example is when F (t) = t and in this case, there is no ambiguity.As pointed out, this gives the Riemann Stieltjes integral provided f is continuous on [a,b]and vanishes near a and b. One must remember that single points can have positive mea-sure in the Stieltjes case where you just have an increasing integrator function which couldhave jumps.

14.12 The Vitali Covering TheoremsThese theorems are remarkable and fantastically useful. I will use m for µF where F (t) = t.This yields Lebesgue measure which gives the measure of an interval to be the length of theinterval. As earlier, B(x,r) will be the interval (x− r,x+ r) . Thus m(B(x,αr)) = αm(x,r)whenever α > 0. Also note that the closure of B(x,r) is the closed interval [x− r,x+ r].This is because the closure simply adds in the limit points which are not in the open intervaland these are the end points. This all extends to many dimensions and so I am using the