If {fn} is subsetof L+, fn decreases pointwise to f, and integral f1 <
infinity, then integral f = lim integral fn.

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Request for Question Clarification by skermit-ga on 13 Nov 2002 10:20 PST

What exactly is the question?

Hi, madukar:

Because of the notation "L+" and some other hints, I will assume that
you are studying the Legesgue definition of the integral.  In my
notation below any integrals stated should be understood as defined in
this way. [I will add a comment or clarification regarding similar
topics under other definitions of the integral.]

This result may be deduced from Lebesgue's Monotone Convergence
Theorem:

http://planetmath.org/encyclopedia/LebesguesMonotoneConvergenceTheorem.html

or more directly from the Dominated Convergence Theorem:

[see Item 4., which includes a proof outline using Fatou's lemma]

http://math.holycross.edu/~dbd/recent_courses/analysis_seminar/assignments/discussion17/node1.html

Here we assume that "L+" denotes the set of nonnegative Lebesgue
integrable functions on the real line R (or more generally, any
measurable subset of R).

We assume that "f_n decreases pointwise to f" means that both {f_n} is
pointwise decreasing and that it has pointwise limit f.

Since the (Lebesgue) integral of nonnegative function f_1 is finite,
the integrals of all the remaining functions in the sequence are also
finite.

Because the deduction of the result above from the Dominated
Convergence Theorem is essentially trivial, I will give a proof that
invokes only the comparatively weaker Monotone Convergence Theorem
(MCT):

THM.  Let {f_n} is a sequence of measurable functions from a
measurable subset E of R into R, and suppose for each n and almost
every x in E:

0 <= f_n(x) <= f_n+1(x)

If there exists a function f from E into R such that:

 f(x) = LIMIT f_n(x) AS n -> +oo for almost every x in E, 

then f is a measurable function and:

 INTEGRAL f(x) dx = LIMIT (INTEGRAL f_n(x) dx) AS n -> +oo

[End of THM. statement]

The big difference between the situation described by the MCT and the
result you ask about is the direction of monotonicity.  Your functions
are monotone decreasing (at all points of the domain) while those in
the MCT are assumed to be monotone increasing (at almost every point
of the domain).

This naturally suggests the following construction.  Consider the
sequence of functions f_1 - f_n.  By our assumption of montonicity,
this is also a nonnegative sequence of measurable functions, whose
pointwise limit is f_1 - f.

Therefore by MCT, we know f_1 - f is a measurable function and:

INTEGRAL (f_1(x) - f(x)) dx 
 = LIMIT (INTEGRAL (f_1(x) - f_n(x)) dx) AS n -> +oo

Now by linearity of the Lesbegue integral and limits, and since:

f(x) = f_1(x) - (f_1(x) - f(x))

we deduce that f is measurable and:

INTEGRAL f(x) dx 
 = INTEGRAL f_1(x) dx - INTEGRAL (f_1(x) - f(x)) dx
 = LIMIT f_1(x) dx - LIMIT (INTEGRAL (f_1(x) - f_n(x)) dx) AS n -> +oo
 = LIMIT [f_1(x) - (f_1(x) - f_n(x))] dx AS n -> +oo
 = LIMIT f_n(x) dx AS n -> +oo

QED

regards, mathtalk-ga

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Request for Answer Clarification by madukar-ga on 14 Nov 2002 12:23 PST

hello, 

  the question is 

 if  {fn}  is contained in L+(L+= the space of all measurable
functions from X to[0,infinity], fn decreases pointwise to f, and
integral f1 < infinity, then integral f = lim integral fn.

 can you explain me why you have used negatilve sign to define the 
function f? what i actually need is integral f = lim integral fn. if
you dont mind could you please clarify my doubt.

  thanks,

  madukar

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Clarification of Answer by mathtalk-ga on 14 Nov 2002 17:53 PST

Hi, madukar:

Let me try to give brief answers to the two points you ask about.

1) Why did I use the negative sign to "define" the function f? 
Actually, f is defined (by your conditions) as the pointwise limit of
the sequence {f_n}.

What I did define was a new sequence of functions, say:

g_n = f_1 - f_n

These functions are also in L+, because since f_n <= f_1 at each
point, the difference g_n is nonnegative (and because the
measurability of g_n follows from that of f_1 and f_n).

Why work with the sequence g_n ?  It is because these functions
satisfy the hypotheses of the Monotone Convergence Theorem.  Where the
sequence f_n is decreasing, the sequence g_n is increasing.  This is a
direct consequence of using the "negative sign" or difference f_1 -
f_n to define g_n.  Notice that the difference involves a fixed
function f_1 minus the sequence of functions f_n.  Therefore as the
functions f_n decrease (toward their limit f), the functions g_n
increase (toward the limit g = f_1 - f).

2) I understand very clearly that the desired conclusion is to show:

integral f = limit integral f_n  [Note: emulating your notation now.]

Note that f = f_1 - (f_1 - f) = f_1 - g, so it is certainly true that
(to restate my final step in your notation):

integral f = integral (f_1 - g) = (integral f_1) - (integral g)

= (integral f_1) - limit (integral g_n)  [Note: This is where we use
MCT.]

= limit (integral f_1 - g_n)  [This is just linearity of limits &
integrals.]

= limit integral f_n

The crucial step is being able to apply MCT to the sequence {g_n} as
it converges "upward" pointwise to g:

integral g = limit integral g_n

Again, we cannot directly apply MCT to the sequence {f_n} because that
sequence is converging "downward" pointwise to f.  By introducing the
new sequence {g_n} we can apply the Monotone Convergence Theorem to
it.

Let's recall the hypotheses of the Monotone Convergence Theorem as
they would apply to {g_n}:

- The functions g_n are measurable and nonnegative, as explained above
in (1).

- The functions g_n converge pointwise upward to g, as also explained
above.

The Monotone Convergence Theorem thus applies; g is a measurable
function and:

integral g = limit integral g_n

Now in general it might be true, with the hypotheses of MCT, that this
limiting integral is (positively) infinite (and the equation is then
+oo = +oo).  But in the more restrictive circumstances of your
problem, we are told that f_1 has a finite integral, ie.

integral f_1 < infinity

As a consequence of this, since 0 <= f <= f_1, the integral of f will
also be finite, as will the integral of g = f_1 - f.

Do these comments help to dispel your doubts?  If not, I encourage you
to go step by step through my original answer, which is arranged as a
linearly ordered proof, and identify any particular steps which seem
doubtful or inexplicable.  The purpose is to prove integral f = limit
integral f_n, and each step in the proof builds toward this goal.

regards, mathtalk-ga

i hope to back with some more questions for you .