Timeline for A Lebesgue integrable function whose absolute value is not Lebesgue integrable
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| May 1, 2016 at 2:01 | vote | accept | telemaco | ||
| Apr 29, 2016 at 18:41 | comment | added | layman | @julian.marr ...converse, that if $u$ and $v$ are integrable, then $f$ is integrable in the absolute value sense. | |
| Apr 29, 2016 at 18:41 | comment | added | layman | @julian.marr One final note: if $f(x) = u(x) + i v(x)$, you might be tempted to say "well, I want to say $f$ is integrable if $u$ and $v$ are, i.e., if we have $\int |u| \,d\mu < \infty$ and $\int |v| \,d\mu < \infty$." And I would reply that you have a great idea, and that actually you could define $f$ being integrable in this way. In fact, it's equivalent to $|f|$ being integrable. I recommend you try to prove that. Prove that if $f(x) = u(x) + i v(x)$, and if $f$ is integrable in the absolute value sense, then it is integrable in the $u$, $v$ integrability sense. Then prove the | |
| Apr 29, 2016 at 18:39 | comment | added | layman | @julian.marr $|a + bi|$ is defined to be $\sqrt{a^{2} + b^{2}}$, which is a real number. So $|f|$ is a real valued function if $f$ is complex valued. So we can talk about $\int |f| \,d\mu$, and this is a real number now, so we can talk about when it's finite. | |
| Apr 29, 2016 at 18:38 | comment | added | layman | @julian.marr Sorry for my late reply. If $f$ is a complex valued function, then we can write $f(x) = u(x) + i v(x)$ for real valued functions $u$ and $v$ ($u$ is called the real part of $f$, and $v$ is called the imaginary part). We define $\int f \,d\mu = \int u \,d\mu + i \int v \,d\mu$. When should this be integrable? The right hand side is a complex number. We don't have an ordering $<$ on the complex numbers to say that a complex number $a + bi < \infty$. So when should we say $\int f \,d\mu$ is integrable? Well, we can take the absolute value of a complex number. | |
| Apr 29, 2016 at 17:22 | history | edited | User8128 | CC BY-SA 3.0 |
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| Apr 29, 2016 at 17:16 | comment | added | Vladimir | One cannot write $\int f d\mu<\infty$ if $\int f d\mu$ is not defined in the first place. On the other hand, if $f$ is $\mu$-measurable, then the integral $\int|f|d\mu$ is always defined (but can be infinite), and if it is finite, then $f$ is integrable and the integral of $f$ is well defined. | |
| Apr 29, 2016 at 17:14 | comment | added | User8128 | Well, since $\lvert f \rvert$ is a non-negative function, we can define $ \int_\Omega \rvert f \lvert d \mu$ without ever thinking about $\int_\Omega f \, d\mu$. By contrast, the definition of $\int_\Omega f \, d\mu$ hinges on the definition of integration of non-negative functions, so perhaps we do this because of the order in which we think about such things. Alternatively, if we would like to consider "integrable functions" as a linear space, the natural norm on that space is $\int_\Omega \lvert f \rvert d\mu$ so it is a natural condition to check that this value is finite. | |
| Apr 29, 2016 at 17:14 | comment | added | telemaco | @user46944 Could you elaborate further on this please? | |
| Apr 29, 2016 at 17:13 | comment | added | layman | @julian.marr What if $f$ is complex valued? That's where absolute value is truly needed. | |
| Apr 29, 2016 at 17:10 | comment | added | telemaco | I always see that a condition for integrability is $\int_\Omega |f|d\mu<\infty$. But this is exactly the same as $\int_\Omega fd\mu<\infty$. Then why the extra "$|\cdot|$"? | |
| Apr 29, 2016 at 17:07 | history | answered | User8128 | CC BY-SA 3.0 |