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This is a rather basic question on parts of the Lebesgue-Integral formula. Intuitively it should be clear, but I am searching for a mathematical explanation.

If $f: (\Omega, \mathcal{A}) \rightarrow (\mathbb{R}, \mathfrak{B}(\mathbb{R}))$ is a measurable function and $\mu$ a measure on $(\Omega, \mathcal{A})$ and $\mu(\Omega) < \infty$, than we can write the Integral of $f$ wrt $\mu$ as $$\int f d\mu = \int f^+ d\mu - \int f^- d\mu =\int_{0}^{\infty} \mu \{y|f(y) \geq x\} dx + \int_{-\infty}^{0} [\mu \{y|f(y) \geq x\} - \mu(\Omega)] dx. $$

I don't get how to derive the second part, i.e. where does the expression $\mu \{y|f(y) \geq x\} - \mu(\Omega)$ come from? (I know that this concerns the potentially negative part of the function, but at the moment I am unable to get, how I derive this second expression). Some help would be awesome!

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  • $\begingroup$ Write $f = f_+ - f_-$. Also, you need to assume $\mu(\Omega) < \infty$ or else the second expression is undefined. $\endgroup$ Dec 20, 2020 at 18:12

2 Answers 2

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Hint: $$ \mu( \{y \quad | \quad f(y) < x \}) + \mu (\{y \quad | \quad f(y) \ge x\} = \mu (\Omega) $$

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  • $\begingroup$ Thank you for the hint! I have now: $$ - \int f^- d\mu = - \int_0^{\infty} \mu( -f \geq x) dx = - \int_0^{\infty} \mu(f < -x) dx = \\ - \int_{-\infty}^0 \mu(f < x) dx = - \int_{- \infty}^0 [\mu(\Omega)- \mu(f \geq x) ]dx = \int_{- \infty}^0 [ \mu(f \geq x)- \mu(\Omega) ]dx$$ Correct? $\endgroup$
    – Galois1763
    Dec 22, 2020 at 12:27
  • $\begingroup$ It's almost correct, you only need to justify the marked equality. $- \int_0^{\infty} \mu( -f \geq x) dx = - \int_0^{\infty} \mu(f \color{red} \leq -x) dx =^{?} \int_0^{\infty} \mu(f \color{red} < -x) dx $ $\endgroup$ Dec 22, 2020 at 13:24
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Hint In addition to @AsemAbdelraouf 's hint, apply Fubini's theorem to the integral \begin{equation} \int_{\mathbb R}\int_{\Omega} \left(\mathbf{1}_{0\le x\le f(y)}-\mathbf{1}_{f(y)<x<0}\right) d x d\mu(y) \end{equation}

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