The Student Room Group

Monotone Convergence



Monotone Convergence Theorem:



Does this follow straight from the theorem?

Have I got to prove something to show it follows from the theorem? If so, what?
Reply 1
It almost follows from the theorem. But first you have to write Inf=Sfn\displaystyle \int_{I_n} f = \int_S f_n for some fnf_n in such a way that (fn)(f_n) is an increasing sequence tending to ff. (Then you have something that satisfies the hypotheses of the theorem.) Can you do that?

Then the case for gg is similar by considering positive and negative parts.
Reply 2
Original post by nuodai
It almost follows from the theorem. But first you have to write Inf=Sfn\displaystyle \int_{I_n} f = \int_S f_n for some fnf_n in such a way that (fn)(f_n) is an increasing sequence tending to ff. (Then you have something that satisfies the hypotheses of the theorem.) Can you do that?

Then the case for gg is similar by considering positive and negative parts.


Let fn=fχInf_n = f \chi_{I_n} where χ\chi is the characteristic function.

Then since f0f \geqslant 0, (fn)(f_n) is an increasing sequence of functions which converges everywhere to ff.

So the result now follows from the MCT. Right?

Any function gg can be written as g=g+gg = g^+ - g^- where g+,g0g^+,g^- \geqslant 0 are the +ve and -ve parts of gg.

Then we know that if gL1(In)g \in L^1 (I_n) then g+,gL1(In)g^+ , g^- \in L^1(I_n).

Ing=In(g+g)=Ing+Ing\displaystyle \int_{I_n} g = \int_{I_n} (g^+ - g^-) = \int_{I_n} g^+ - \int_{I_n} g^-

If I let gn+=g+χIng_n^+ = g^+\chi_{I_n} then (gn+)(g_n^+) is increasing and converging everywhere to g+g^+ and let gn=gχIng^-_n = g^- \chi_{I_n} so (gn)(g_n^-) is an increasing sequence of functions which converges everywhere to gg^-.

So Ing=Sgn+Sgn\displaystyle \int_{I_n} g = \int_S g_n^+ - \int_S g_n^- and the results for gn+g_n^+ and gng_n^- follow from the MCT.

What is the significance of the question saying Ing\int_{I_n} |g| converge?

Does the result for gg now follow from the MCT?
(edited 11 years ago)
Reply 3
Original post by TheEd
Let fn=fχInf_n = f \chi_{I_n} where χ\chi is the characteristic function.

Then since f0f \geqslant 0, (fn)(f_n) is an increasing sequence of functions which converges everywhere to ff.

So the result now follows from the MCT. Right?
That's right.

Original post by TheEd
Any function gg can be written as g=g+gg = g^+ - g^- where g+,g0g^+,g^- \geqslant 0 are the +ve and -ve parts of gg.

Then we know that if gL1(In)g \in L^1 (I_n) then g+,gL1(In)g^+ , g^- \in L^1(I_n).

Ing=In(g+g)=Ing+Ing\displaystyle \int_{I_n} g = \int_{I_n} (g^+ - g^-) = \int_{I_n} g^+ - \int_{I_n} g^-
You could essentially have stopped here. Since g+,gg^+,g^- are nonnegative, you could have just said that the result follows from the first part of the question.

Original post by TheEd
What is the significance of the question saying Ing\int_{I_n} |g| converge?


That's just the statement that gχIng\chi_{I_n} is integrable for each nn. It ensures that Ing+\displaystyle \int_{I_n} g^+ and Ing\displaystyle \int_{I_n} g^- are all finite so that you're not taking infinity away from infinity or anything horrible.
Reply 4
Original post by nuodai
That's right.


OK. So how do I exactly structure the argument?

Let fn=fχInf_n = f \chi_{I_n}.

Then since f0f \geqslant 0 we have that (fn)(f_n) is an increasing sequence of functions which converges everywhere to ff.

Now we are supposing Inf\int_{I_n} f converges.

So by MCT, Inf=limnInfn\int_{I_n} f = \lim_{n\to\infty} \int_{I_n} f_n

How does the results that fL1(S)f \in L^1 (S) and Sf=limnInf\int_S f = \lim_{n\to\infty} \int_{I_n}f now follow from the MCT?
(edited 11 years ago)
Reply 5
Original post by TheEd
OK. So how do I exactly structure the argument?

Let fn=fχInf_n = f \chi_{I_n}.

Then since f0f \geqslant 0 we have that (fn)(f_n) is an increasing sequence of functions which converges everywhere to ff.

Now we are supposing Inf\int_{I_n} f converges.

How does the results that fL1(S)f \in L^1 (S) and Sf=limnInf\int_S f = \lim_{n\to\infty} \int_{I_n}f now follow from the MCT?


Let fn=fχInf_n = f \chi_{I_n} where f0f \ge 0. Then fnL1(S)f_n \in L^1(S) since fL1(In)f \in L^1(I_n).

Then (fn)(f_n) is an increasing sequence in L1(S)L^1(S) tending to ff

So by the monotone convergence theorem

Inf=SfnSf\displaystyle \int_{I_n} f = \int_S f_n \to \int_S f and fL1(S)f \in L^1(S)
(edited 11 years ago)
Reply 6
Original post by nuodai
Let fn=fχInf_n = f \chi_{I_n} where f0f \ge 0. Then fnL1(S)f_n \in L^1(S) since fL1(In)f \in L^1(I_n).

Then (fn)(f_n) is an increasing sequence in L1(S)L^1(S) tending to ff

So by the monotone convergence theorem

Inf=SfnSf\displaystyle \int_{I_n} f = \int_S f_n \to \int_S f and fL1(S)f \in L^1(S)


OK. So why do we have to assume that the integral of the absolute value of gg converges for the 2nd part?

Ing=Ing+g=Sgn+gn=...\displaystyle \int_{I_n} |g| = \int_{I_n} |g^+ - g^-| = \int_S |g_n^+ - g_n^-| = ...

Can I split it into 2 integrals like I did before with |\cdot | there?
(edited 11 years ago)
Reply 7
Original post by TheEd
OK. So why do we have to assume that the integral of the absolute value of gg converges for the 2nd part?

Ing=Ing+g=...\displaystyle \int_{I_n} |g| = \int_{I_n} |g^+ - g^-| = ...

Can I split it into 2 like I did before with |.| there?


g=g++g\left| g \right| = g^+ + g^-, and g+,g0g^+,g^- \ge 0, so if Ing(=Ing++Ing)<\displaystyle \int_{I_n} \left| g \right| \left(= \int_{I_n} g^+ + \int_{I_n} g^-\right) < \infty then Ing+<\displaystyle \int_{I_n} g^+ < \infty and Ing<\displaystyle \int_{I_n} g^- < \infty, so that the difference Ing+Ing\displaystyle \int_{I_n} g^+ - \int_{I_n} g^- is well-defined. This is something you should have come across in the definition of (μ\mu-)integrability.
(edited 11 years ago)

Quick Reply

Latest