Convergence of integrals of uniformly convergent functions

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. 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?

I think this is a beautiful exercise, but I got disappointed. Lebesgue
techniques usually make it much simpler to prove theorems about
Riemann integrals, but in this case the standard proof, based on
partitions and upper and lower sums, is much simpler, at least than
the proof I could come up with.

Thank you

Amanda.

.

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