I would like to show that $\int_{E\cup F}f=\int_E f+\int _F f$, where $E\cap F=\emptyset$ and $E,F$ are Lebesgue measurable sets.


First I tried to show that in general I can write $\int (f+g)=\int f+\int g$ like so:


Suppose that there are two sequences $(\varphi_n)$ and $(\psi_n)$ such that $\lim_{n\to\infty}\varphi_n=f$ and $\lim_{n\to\infty}\psi_n=g$, where these sequences are made up of non-negative, integrable, simple functions. Then applying Lebesgue's Monotone Convergence Theorem (that is, if we have $f_i$ non-negative and measurable: $f_1\leq\ldots\leq f_n\leq\ldots$, then $\lim_{n\to\infty}\int f_n=\int\lim_{n\to\infty}f_n$), we can get the following:

\begin{align} &\int (f+g)\\ &=\lim_{n\to\infty} \int(\varphi_n+\psi_n)\\ &=\lim_{n\to\infty}\int \varphi_n+\lim_{n\to\infty}\int \psi_n\\ &=\int f+\int g \end{align}

So I (think) I've proved that $\int_T (f+g)=\int_T f+\int_T g$, but this is for a common domain $T$. I'm not sure how to extend this result to split the integral with different domains. (Perhaps there is an easier way to show this instead?)

Any help would be appreciated. Thanks.

  • 3
    $\begingroup$ Hint: with $A \subset X$, $X$ a measure space and $A$ measurable, use $\int_A f = \int_X f\cdot\chi_A$, where $\chi_A$ is the indicator function of $A$. Applied to $A = E\cup F$, this will allow you to use the linearity of the integral. $\endgroup$ – Gyu Eun Lee Mar 20 '15 at 4:37

Consider any simple function, $\phi = \sum a_k\chi_{E_k}$. Then, \begin{align} \int_{E\cup F} \phi = \int\phi\cdot\chi_{E\cup F} &= \sum_{k=1}^na_k\mu(E_k\cap(E\cup F))\\ &= \sum_{k=1}^na_k\mu(E_k\cap E) +\sum_{k=1}^na_k\mu(E_k\cap F)\\ &= \int_{E} \phi + \int_{F} \phi. \end{align} Can you continue from here?

  • $\begingroup$ @ hjhjhj57: I think I've got it. Supposing that $E\cap F=\emptyset$, we know that $\mu(E\cup F)=\mu(E)+\mu(F)$. Then, $\int_{E\cup F} f= \int f\cdot \chi_{E\cup F}=\int f\cdot \mu(E\cup F)=\int f\cdot (\mu(E)+\mu(F))=\int (f\cdot \mu(E) +f\cdot \mu(F))=\int f\cdot\mu(E)+\int f\cdot\mu(F)=\int f\cdot \chi_E+\int f\cdot \chi _F=\int _E f+\int_F f$ $\endgroup$ – Sujaan Kunalan Mar 20 '15 at 16:05
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    $\begingroup$ Actually what I said was wrong. It should be: $\int_{E\cup F}f=\int f\chi_{E\cup F}=\int f\chi_E+\int f\chi_F=\int_E f+ \int_F f$. Thanks for the hint. $\endgroup$ – Sujaan Kunalan Mar 20 '15 at 18:38

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