Properties of L1, lebesgue integreable functions

[onthetopo]

Homework Statement

(X,S,u) is measur space, f,g are in L1. Prove that:
1.min(f,g) is in L1
2.$$min(\int(fdu),\int(gdu))\geq\int(min(f,g)du)$$
3. when does equality hold?

The attempt at a solution
1.Since both f,g are in L1, and min must be one of the f or g, both of which are in L1, thus min (f,g) is in L1.
2.Star with simple functions and extend to limit?
3. Guess: equality hold if f=g

 

[Office_Shredder]

Question 1 isn't asking if min(f,g) is in L1... actually, I'm not really sure what it's asking, since you didn't post a function. was it supposed to be min(f,g)?

For 2, notice that f>=min(f,g) so
$$\int fdu \geq \int min(f,g) du$$

and the same for g.

For 3, equality holds under a much broader class of conditions... try drawing some functions on R like x, x², 8, and sin(x) and try some comparisons. Or you can try to glean it from the proof of 2

 

[onthetopo]

sorry about that, I've corrected question 1.

For3. Are you saying that monotonic functions satisfy equality?

 

[Office_Shredder]

Only if sin(x) is monotonic :p

Try drawing the graphs y=sin(x) and y=8 and see if it satisfies equality. Then try the same for y=sin(x) and y=1/2. What's the difference?

 

[rochfor1]

#1 Is a little more complicated than your proof, because which function is the minimum may change from point to point. Think about f(x)=x and g(x)=x^2.

 

[onthetopo]

Only if sin(x) is monotonic :p

Try drawing the graphs y=sin(x) and y=8 and see if it satisfies equality. Then try the same for y=sin(x) and y=1/2. What's the difference?

I see, so the equality holds when there doesn't exist any x such that f(x)=g(x). , ie, they never cross, but how show rigorously that this is the condition?Regarding question 1. I think the proof is a one-liner using the comparison theorem, since if we can show|min(f,g)|<|f|, then since f is in L1, by comparison min(f,g) must also be in L1. The only problem is that in order to show |min(f,g)|<|f|, don't we need both f and g to be nonnegative?

 

[rochfor1]

Yes, for that you would need f, g >= 0. Try this: let $$E = \{ x | f(x) \leq g(x) \}$$,
so
$$\begin{align*}<br /> \int |\min ( f, g )| d\mu = \int_E |\min ( f, g )| d\mu + \int_{ E^c } |\min ( f, g )| d\mu = \int_E |f| d\mu + \int_{ E^c } |g| d\mu,<br /> \end{align*}$$
and see where you can go from there.

 

[onthetopo]

Yes, for that you would need f, g >= 0. Try this: let $$E = \{ x | f(x) \leq g(x) \}$$,
so
$$\begin{align*}<br /> \int |\min ( f, g )| d\mu = \int_E |\min ( f, g )| d\mu + \int_{ E^c } |\min ( f, g )| d\mu = \int_E |f| d\mu + \int_{ E^c } |g| d\mu,<br /> \end{align*}$$
and see where you can go from there.

E is a measurable set, let x be the characteristic function, then
|f|xE is in L1
|g|xEc is in L1
and |f|xE+|g|xEc=|min(f,g)| is in L1
thus min (f,g) is in L1
QED
please tell me this is right, exam is next week, thanks

 

[rochfor1]

That's exactly right!