Re: Convergence of integrals of uniformly convergent functions

On 28-03-2007 22:10, Amanda wrote:

I'm studying measure theory and in an exercise we were asked to use
Lebesgue's techniques to prove the following theorem about Riemann
integrals:

If f_n is a sequence of real valued Riemann integrable function
defined on a compact interval I and converges uniformly to a function
f, then f is Riemann integrable and Int f(x) dx = lim Int f_n(x) dx
(integrals over I = [a, b]).

At similar result for Lebesgue integrals had been proved, so we could
use it. The proof I came up with is:

Since the functions f_n are Riemann integrable, then they are Lebesgue
integrable with respect to the Lebesgue measure u and th 2 integrals
coincide. Since f_n converges uniformly to f and u(I) = b - a < oo,
it follows f is Lebesgue integrable over I and

lim Int f_n du = Int f du, all integrals being computed over I.
Therefore,

(Riemann) lim Inf f_n(x) dx = Int f du (lebesgue). This equality
shows that, if we now prove f is Riemann integrable over I, than we
are done.

Since the functions f_n are Riemann integrable, they are bounded and
the set D_n of the their discontinuities in I has u(D_n) = 0. Uniform
convergence of f_n -> f implies uniform boundedness, so that f is
bounded by the same upper bound.

I think that that's a bit fast. What happens is that each f_n is bounded
(by definition of Riemann integrable) and that the uniform limit of a
sequence of bounded functions is bounded.

So, to complete the proof, it remains
to show u(D) = 0, where D is the set of the discontinuities of f in
I,

In virtue of uniform convergence, if, for some x in I, continuity of
the functions f_n at x occur for infinitely many n, then f is
continuous at x, because then f is the uniform limit of a subsequence
of functions continuous at x. It follows D is subset of the set of
elements of I that, with possibly exception of a finite number of
values of n, belong to all sets D_n, that is liminf D_n. Since liminf
D_n is subset of Union D_n, the subadditivity and monotonicity of the
measure implies that

u(D) <= u(liminf D_n) <= u(Union D_n) <= Sum u(D_n) =0. which shows f
is continuous almost everywhere on I. It follows f is Riemann
integrable and we are done.
Is there any flaw in this proof?

No, but it is too much complicated. You could just have said that, since
_f_ must be continuous at each point _x_ at which each f_n is
continuous, therefore D is a subset of union D_n.

Best regards,

Jose Carlos Santos
.

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