I've had a look to see whether this already exists but can't find it.

Context: Measure theory, specifically in discussion about the Lebesgue measure of a positive function.

Some sources (some would say the "conventional" ones) define the negative part of a function as: $$\forall x \in S: f^{-} (x) = -\min (0, f(x))$$ that is, inverting it and turning it into a positive function.

One is then able to write: $$\int f (x) = \int f^{+} (x) - \int f^{-} (x)$$

But I have before me Arne Broman's Introduction to Partial Differential Equations (1970) which defines the concept of a negative part as:

$$\forall x \in S: f^{-} (x) = \min (0, f(x))$$

thus leaving the negative part as, indeed, a "negative function".

This leads us to:

$$\int f (x) = \int f^{+} (x) + \int f^{-} (x)$$

You can then perform your analysis in exactly the same way as Lebesgue, but you just have to be careful that you haven't got the sign round the wrong way.

What is the current view on this? Is it more helpful to do it the first way or the second? I fully expect that once you've learned it one way you'd be resistant to pressure to go and have to rethink it for little (if any) gain -- but is there acceptance among the newer generations to accept this tentative rethink of the paradigm?

Or is it a regional thing, where different schools use different paradigms?


1 Answer 1


Every integral of an integrable function can be split into two parts. The one is the integral over the set {x:f(x)$\geq $0} label it A+ and the set {x:f(x)<0} label it A- therefore the integral is split in$\int_{A+} f(x)dx +\int _{A-}f(x)dx$ But on A+ we have f(x)=max{0,f(x)} and on A- we have f(x)=min{0,f(x)}. Defining f+(x)=max{0,f(x)} and f-(x)=min{0,f(x)} we get $\int_{A+} f(x)dx$+$\int_{A-} f(x)dx$ =$\int f(x)dx=\int f_{+}(x)dx+\int f_{-}(x)dx$

  • $\begingroup$ Well basically yes, but this does not answer the question. $\endgroup$ Jun 18, 2022 at 14:38
  • $\begingroup$ Don't waste much time about this! It is a matter of definition! The definition of all classical books (Royden, Kantorovich, Dunford etc) define it in the way I described in my answer! $\endgroup$
    – user1054388
    Jun 18, 2022 at 17:19
  • $\begingroup$ Would you care to improve it? For example: grammar, consistent $\LaTeX$, improved layout -- then it will be easier to understand what you mean to say. $\endgroup$ Jun 18, 2022 at 17:49

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