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If $f$ and $g$ are nonnegative Lebesgue measurable functions, then we know that $\int (f+g) d\lambda = \int f d \lambda + \int g d \lambda $. Given the difinition of integral of an arbitrary Lebesgue measurable function, that is $ \int f d\lambda = \int f^+ d\lambda - \int f^- d\lambda$, how do you prove $\int (f_1 - f_2) d\lambda = \int f_1 d\lambda - \int f_2 d\lambda $ for nonnegative Lebesgue integrable functions $f_1$ and $f_2$ ?

That $f_1 - f_2$ is Lebesgue integrable is easily seen. Now, I can show that $\int (-f) d\lambda = -\int f d\lambda$, as $\int (-f) d\lambda = \int (-f)^+ d\lambda - \int (-f)^- d\lambda = \int f^- d\lambda - \int f^+ d\lambda = -\int fd\lambda$. Any ideas how to proceed?

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So I came up with a way to do it:

The idea is to split the domain of integration into part on which $f_1 - f_2$ is positive and part on which it is negative, and use the identity

$$ \int |f_1 - f_2| = \int (f_1 -f_2)^+ + \int (f_1-f_2)^- $$

More precisely we let $E = \{x: f_1(x) \ge f_2(x) \}$, so $\int_E |f_1 - f_2| = \int_E (f_1 - f_2) $. Now since $f_1 - f_2$ is nonnegative we can apply the argument of drhap to deduce that $\int_E (f_1 - f_2) = \int_E f_1 - \int_E f_2$. Similarly, $\int_{E^c} |f_1 - f_2| = \int_{E^c} -(f_1-f_2) = \int_{E^c} f_2 -\int_{E^c}f_1$. From what I already proved, the negative sign can be taken out of the second integral. Now if we combine these we'll get the result.

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  • $\begingroup$ Looks okay to me. $\endgroup$ – drhab Apr 28 '14 at 17:32
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Let $h=f+g$ and notice that $$h^+-h^-=f^+-f^-+g^+-g^-.$$ This means $$h^++f^-+g^- = h^-+f^++g^+,$$ and since both sides have constant sign, $$\int h^++ \int f^-+ \int g^- = \int h^-+ \int f^++ \int g^+.$$ Now collect terms in a suitable way.

This proof is taken from W. Rudin's Real and complex analysis.

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  • $\begingroup$ That didn't work. The last line give me $\int h =\int f + \int g$, not what I want to prove. $\endgroup$ – ywx Apr 28 '14 at 15:02
  • $\begingroup$ @ywx On base of $\int\left(f+g\right)=\int f+\int g$ you can finish the job by noting that $\int\left(f_{1}-f_{2}\right)+\int f_{2}=\int\left[\left(f_{1}-f_{2}\right)+f_{2}\right]=\int f_{1}$ and consequently $\int\left(f_{1}-f_{2}\right)=\int f_{1}-\int f_{2}$ $\endgroup$ – drhab Apr 28 '14 at 15:36
  • $\begingroup$ As I am following the book, the property $\int (f + g) = \int f +\int g$ hasn't yet been generalized to arbitrary functions, I can only assume it's true for nonnegative functions. We can try splitting the domain of integration to positive and negative part of $f_1 - f_2$ $\endgroup$ – ywx Apr 28 '14 at 16:13
  • $\begingroup$ I think I've got an answer however SE won't let me answer (<10 rep) That's frustrating $\endgroup$ – ywx Apr 28 '14 at 16:50
  • $\begingroup$ Now give your answer. (>10 rep) $\endgroup$ – drhab Apr 28 '14 at 17:00
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Use the usual trick... split up the cases where $f_1-f_2$ is positive and negative and use your original result, then combine terms properly.

Namely, consider $(f_1-f_2)^+$ and $(f_1-f_2)^-$ now what exactly is

$$\int (f_1-f_2)^+ - \int(f_1-f_2)^-$$

?

Now use linearity etc.

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  • $\begingroup$ "Now use linearity etc." I can't, this is what I want to prove. $\endgroup$ – ywx Apr 28 '14 at 15:04

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