2
$\begingroup$

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.

Attempt:

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

Proof.

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.

$\endgroup$
  • 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
4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.