I was getting back into this for something the other day.Is it true that if h = f + g (pointwise) and h is Lebesque integrable then:int h = int f + int g (under the lebesque itegral)? each is defined and the sum is always correct, right?

Last edited by CactusMan on January 16th, 2015, 11:00 pm, edited 1 time in total.

- MiloRambaldi
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QuoteOriginally posted by: CactusManI was getting back into this for something the other day.Is it true that if h = f + g (pointwise) and h is Lebesque integrable then:int h = int f + int g (under the lebesque itegral)? each is defined and the sum is always correct, right?A reasonable question, but in the wrong forum.

- Cuchulainn
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QuoteOriginally posted by: MiloRambaldiQuoteOriginally posted by: CactusManI was getting back into this for something the other day.Is it true that if h = f + g (pointwise) and h is Lebesque integrable then:int h = int f + int g (under the lebesque itegral)? each is defined and the sum is always correct, right?A reasonable question, but in the wrong forum.Indeed.BTW The Radon stuff was moved to Student by request.Quoteeach is defined In what sense? Depending on the answer to this question, you may really be referring to a good old Riemann integral and LebesGue is redundant.

Last edited by Cuchulainn on January 16th, 2015, 11:00 pm, edited 1 time in total.

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- MiloRambaldi
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QuoteOriginally posted by: CactusManI was getting back into this for something the other day.Is it true that if h = f + g (pointwise) and h is Lebesque integrable then:int h = int f + int g (under the lebesque itegral)? each is defined and the sum is always correct, right?The simple answer is no. Let [$]A\subseteq[0,1][$] be a non-measurable set, and [$]h=1_{[0,1]}[$], [$]f=1_{A}[$] and [$]g=1_{[0,1]\setminus A}[$] (indicator functions).Then h = f + g and h is Lebesgue integrable with integral 1, but neither f nor g is Lebesgue measurable, and thus their Lebesgue integrals are not defined.

QuoteOriginally posted by: MiloRambaldiQuoteOriginally posted by: CactusManI was getting back into this for something the other day.Is it true that if h = f + g (pointwise) and h is Lebesque integrable then:int h = int f + int g (under the lebesque itegral)? each is defined and the sum is always correct, right?The simple answer is no. Let [$]A\subseteq[0,1][$] be a non-measurable set, and [$]h=1_{[0,1]}[$], [$]f=1_{A}[$] and [$]g=1_{[0,1]\setminus A}[$] (indicator functions).Then h = f + g and h is Lebesgue integrable with integral 1, but neither f nor g is Lebesgue measurable, and thus their Lebesgue integrals are not defined.OK, that is great. that is the breakdown I was trying to remember. I had not thought of this stuff in a long time.Good work. thanks. I am going to send my army of 10 year old Canadian girls in ballet tights to feed you strawberries and rub your feet all day to repay you.

QuoteOriginally posted by: MiloRambaldiQuoteOriginally posted by: CactusManI was getting back into this for something the other day.Is it true that if h = f + g (pointwise) and h is Lebesque integrable then:int h = int f + int g (under the lebesque itegral)? each is defined and the sum is always correct, right?A reasonable question, but in the wrong forum.oh, i am sorry. could someone effect a move. sorry about that.

- MiloRambaldi
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QuoteOriginally posted by: MiloRambaldiQuoteOriginally posted by: CactusManI was getting back into this for something the other day.Is it true that if h = f + g (pointwise) and h is Lebesque integrable then:int h = int f + int g (under the lebesque itegral)? each is defined and the sum is always correct, right?The simple answer is no. Let [$]A\subseteq[0,1][$] be a non-measurable set, and [$]h=1_{[0,1]}[$], [$]f=1_{A}[$] and [$]g=1_{[0,1]\setminus A}[$] (indicator functions).Then h = f + g and h is Lebesgue integrable with integral 1, but neither f nor g is Lebesgue measurable, and thus their Lebesgue integrals are not defined.Actually, that was a poor counterexample. There are models of mathematics, without full choice functions, where every set of real numbers is Lebesgue measurable. (A big discovery in the 1960's.)As suggested in a prior post your question is not really about measure. For a better counterexample let[$]f(x)=\frac1x[$] ([$]x\ne0[$]) and [$]f(0)=0[$]. Then 0 = f - f pointwise, but [$]\infty-\infty[$] is not defined.

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