Re: Proving a sequence of lLebesgue lntegrals converges

On Wed, 07 Nov 2007 10:15:15 +0000, José Carlos Santos
<jcsantos@xxxxxxxx> wrote:

On 07-11-2007 9:43, Amanda wrote:

This is an exercise and I'll show my work. A guy
who knows much more math than me didn't like my
proof, but I don't see anything wrong with it. Maybe
someone here can give me a clue.

Let (X, M, m) be a mesure space and let (f_n) be
a sequence of integrable complex valued functions
defined on X that converges uniformly to a function
f. Show that, if m(X) < oo, then f is integrable an

lim Int f_n dm = Int f dm (integrals always
considered over X)
0
Show this conclusion may fail if m(X) = oo.

My work:

Fix some eps >0. Since f_n --> f uniformly, we can
find a positive integer k (Cauchy Criterion) such
that n >= k => |f_n - f_k| < eps => |f_n| < |f_k| +
eps, which shows the k-tail of (f_n) is dominated by
the function |f_k| + eps (I mean, the function x -->
|f_k(x)| + eps, x in X). Since f_k, by assumption,
is integrable, so is |f_k|. And since m(X) < oo, it
follows that

Int (|f_k| + eps) dm = Int |f_k| dm + Int eps dm =
Int |f_k| dm + eps m(X) < oo,
since both parcels in the right hand side are
finite. Hence, |f_k| + eps is integrable and
dominates the k tail of (f_n). Since f_n --> f, so
does its k-tail, so that we can apply the Dominated
Convergence Theorem to this k-tail to conclude that
f is integrable and

lim (n >= k) Int f_n dm = Int f dm

On the left hand side we just have the limit of
the k tail of (Int f_n dm). Since this k-tail
converges to Int f dm, so does the whole sequence,
and we simply have

lim Int f_n dm = Int f dm , completing the proof.

To see the condition m(X) < oo is actually
essential, put X = [0, oo), M = Lebesgue
sigma-algebra on X and m = Lebesgue measure, so that
m(X) = Length([0, oo)) = oo. For each n=1,2,3...put


f_n(x) = c_n(x)/n, where c_n is the characteristic
function of [0, n]

It's easy to see that f_n converges uniformly on X
to the identically zero function f = 0 (given eps >
0, it suffices to choose n > 1/eps to get 0 <
f_n(x) < eps for all x in X)). So, Int f dm = 0.
But, for each n, we have

Int f_n dm = Int (over [0, n]) f_n dm = n/n =1,
so that

lim Int f_n dm = 1 > 0 = Inf f dm, showing the
condition m(X) < oo is really essential for the
theorem to be true.

In this example, I think it's interesting to point
out that, if some function g dominates (f_n), then
we readily see that

Int g dm > = 1 + 1/2 +1/3.....= oo, so that no
integrable function dominates (f_n), and,
therefore, the proof I gave doesn't work any more.

Yes, but that doesn't prove that the conclusion may fail if m(X) = oo.
That's rather easy; just take f_n(x) = 1/n if x is in [-n,n] and
f_n(x) = 0 otherwise.

??? That's almost exactly what she did. Look up a few paragraphs.

This was my work. Maybe there's a better proof.

Indeed. Take _n_ such that |f(x) - f_n(x)| < 1. Then
|f(x)| < |f_n(x)| + 1 and therefore, by the dominated convergence
theorem, _f_ is integrable. Now, take r > 0 and take a natural _p_
such that |f/x)- f_n(x)| < r/m(X) whenever x is in X and n >= p.
Then, for n >= p,

|int f(x) dx - int f_n(x) dx| = |int (f(x) - f_n(x)) dx|

<= int |f(x) - f_n(x)| dx

< int (r/m(X)) dx

= r.

Best regards,

Jose Carlos Santos


************************

David C. Ullrich
.

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