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Suppose that $(X, M, \mu)$ is a measure space and that $f:X\to \overline{\mathbb R}$ is a measurable function. Let $E\in M$. Is the following true? $$ \int f = \int f \, 1_E + \int f \, 1_{E^c} $$ where $1_E$ represents indicator function.

It $f$ is non-negative valued, then I know that this result is true. It follows by monotone convergence theorem. The result is true even if $f$ is integrable. But I'm not sure if it holds in case of any measurable $f$. $$ \int f = \int \bigl( f \, 1_E + f \, 1_{E^c} \bigr) \overbrace {=}^{?} \int f \, 1_E + \int f \, 1_{E^c}. $$

Is the second equality true? If yes, then how does one prove it and when is it true in general (apart from the cases stated above)?

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  • $\begingroup$ These are the situation where things go wrong: (1) $\int_Ef_+=\infty_Ef_-=\infty$ or (2) $\int_{E^c}f_+=\int_{E^c}f_-=\infty$, or (3) $\int_E f=\pm\infty$ and $\int_{E^c}f=\mp\infty$ $\endgroup$
    – Mittens
    Commented Oct 8, 2022 at 20:11
  • $\begingroup$ @OliverDíaz: I got that now. Thanks a lot. Basically, we want to avoid $\infty -\infty$. $\int f1_A=\int (f1_A)^+-\int (f1_A)^-=\int f^+1_A-\int f^-1_A.$ $\endgroup$
    – Koro
    Commented Oct 8, 2022 at 21:24

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To begin with, let us prove the linearity of the Lebesgue Integral for nonnegative measurable functions based on the MCT. I am going to assume that you have already proved the linearity of the Lebesgue Integral for nonnegative simple functions.

Let $(f_{n})_{n\in\mathbb{N}}$ be an increasing sequence of simple nonnegative functions which converges pointwise to $f$ and $(g_{n})_{n\in\mathbb{N}}$ be an increasing sequence of simple nonnegative functions which converges pointwise to $g$. Consequently, according to the Monotone Convergence Theorem, one gets that: \begin{align*} \int_{\Omega}(f + g)\mathrm{d}\mu & = \lim_{n\to\infty}\int(f_{n} + g_{n})\mathrm{d}\mu\\\\ & = \lim_{n\to\infty}\int_{\Omega}f_{n}\mathrm{d}\mu + \lim_{n\to\infty}\int_{\Omega}g_{n}\mathrm{d}\mu\\\\ & = \int_{\Omega}f\mathrm{d}\mu + \int_{\Omega}g\mathrm{d}\mu \end{align*}

Now we are able to tackle the proposed exercise. In order to do so, notice the linearity applies to any measurable functions $f$ and $g$. With the purpose to conclude so, notice that \begin{align*} (f + g)^{+} - (f + g)^{-} = f + g = (f^{+} - f^{-}) + (g^{+} - g^{-}) \end{align*} Hence we may deduce as well that \begin{align*} (f + g)^{+} + f^{-} + g^{-} = (f + g)^{-} + f^{+} + g^{+}. \end{align*}

Due to the additivity of the Lebesgue integral, one concludes that

\begin{align*} \int_{\Omega}(f + g)^{+}\mathrm{d}\mu + \int_{\Omega}f^{-}\mathrm{d}\mu + \int_{\Omega}g^{-}\mathrm{d}\mu = \int_{\Omega}(f + g)^{-}\mathrm{d}\mu + \int_{\Omega}f^{+}\mathrm{d}\mu + \int_{\Omega}g^{+}\mathrm{d}\mu. \end{align*} Therefore we arrive at the identity: \begin{align*} \int_{\Omega}(f + g)^{+}\mathrm{d}\mu - \int_{\Omega}(f + g)^{-}\mathrm{d}\mu = \int_{\Omega}f^{+}\mathrm{d}\mu - \int_{\Omega}f^{-}\mathrm{d}\mu + \int_{\Omega}g^{+}\mathrm{d}\mu +\int_{\Omega}g^{-}\mathrm{d}\mu \end{align*} and we are done.

Now, let us notice that $\Omega = A\cup A^{c}$, where $A\cap A^{c} = \varnothing$.

Hence $1 = 1_{\Omega} = 1_{A\cup A^{c}} = 1_{A} + 1_{A^{c}}$, whence we get: \begin{align*} \int_{\Omega}f\mathrm{d}\mu & = \int_{\Omega}f1_{\Omega}\mathrm{d}\mu\\ & = \int_{\Omega}f(1_{A} + 1_{A^{c}})\mathrm{d}\mu\\ & = \int_{\Omega}(f1_{A} + f1_{A^{c}})\mathrm{d}\mu\\ & = \int_{\Omega}f1_{A}\mathrm{d}\mu + \int_{\Omega}f1_{A^{c}}\mathrm{d}\mu\\ & = \int_{A}f\mathrm{d}\mu + \int_{A^{c}}f\mathrm{d}\mu \end{align*}

Hopefully this helps!

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  • $\begingroup$ Thanks for the answer. As I mentioned in my post, I know that the result is true for any non negative measurable function (by monotone convergence theorem as you have also shown) and for integrable functions too. But the case, which you left at the end is what's problematic :(. $\endgroup$
    – Koro
    Commented Oct 8, 2022 at 19:36
  • $\begingroup$ @Koro Can you tell me precisely where did you get stuck? $\endgroup$ Commented Oct 8, 2022 at 19:39
  • $\begingroup$ Let me continue from where you left: $\int f=\int f^+-\int f^-=\int f^+1_E+\int f^+1_{E^c}-\int f^-1_{E}-\int f^{-1}1_{E^c}=(\int f^+1_E-\int f^-1_{E})+( \int f^+1_{E^c}-\int f^{-1}1_{E^c})$. The terms inside the bracket in RHS: Is it true that $\int f^+1_E-\int f^-1_{E}=\int f1_{E}$? $\endgroup$
    – Koro
    Commented Oct 8, 2022 at 19:42
  • $\begingroup$ @Koro You can prove such relation for $f + g$ above. Then the proof reduces to what I have proposed. I will edit my answer. $\endgroup$ Commented Oct 8, 2022 at 19:45

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