I need to understand the first part of the solution for this proof, please:

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I don't understand anything about the first sentence of this solution. (if you're interested in the rest of the proof, it can be found here - http://www.math.mcgill.ca/toth/355a2solutions.pdf)

Why is $f=f^{+}-f^{-}$ and what does it mean for the lebesgue integral to be linear (or rather, I know what linear means but how do we know it is linear?) and why does that imply the $f(x) \geq 0$, and why does THAT imply that $g(x)>0$ and why do we care anyway? Can someone please explain as much as they can about this solution (I don't really care about the arithmetic in the full proof, but I don't understand these assumptions -- their necessity or how we know they are true). Thank you!


$f_+(x)=\frac{f(x)+|f(x)|}{2}$ and $f_-(x)=\frac{-f(x)+|f(x)|}{2}$, they are the positive and negative part of the function $f(x)$ respectively. In general if you must demonstrate a theorem like this one, about a function $f(x)$ that it can be positive and/or negative, you usually write $f=f_+-f_-$. If we need to do an integration with $f(x)$ in the demonstration, then we can write:$\int f=\int (f_+-f_-)=$(by linearity of the integral)=$\int f_+ -\int f_-$ so if we demonstrate a result with $\int f_+$ the same demonstration will work for $\int f_-$.

If $f(x)\ge0$ and $0\le t\le b\ge 0$, the function inside the integral is positive so $g(x)\ge 0$ (the integral of a positive function is positive).

  • $\begingroup$ Thanks MattG88....I understand your second sentence,but not those equations in the first line. Is that really what f+ means? What is "f+" or "f-" called using english, so I can search online? And what does it have to do with linearity? $\endgroup$
    – PBJ
    Nov 21 '16 at 0:55
  • $\begingroup$ The positive and negative part of a function select, as the name say, only the positive or negative portion of the function...you can see here anyway en.m.wikipedia.org/wiki/Positive_and_negative_parts $\endgroup$
    – MattG88
    Nov 21 '16 at 1:02
  • $\begingroup$ The first line is to show that it is enough to prove the statement for positive functions. You might ask why? This is because any function can be written as a difference of positive functions. (This answer shows you an explicit way of writing a function f as difference of positive functions f+ and f-). $\endgroup$ Nov 21 '16 at 1:05
  • $\begingroup$ Oh, ok, that was helpful. I think I get it now. Thanks $\endgroup$
    – PBJ
    Nov 21 '16 at 1:19

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