Timeline for Integral means vs infinite convex combinations
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 25 at 16:02 | comment | added | Pietro Majer | In fact if for a measure $\mu$ on [0,1] we denote $F_\mu$ the distribution function $t\mapsto \mu([t,1])$, then the above $f$ is $x\mapsto F_{\delta_x}=\mathbf1_{[0,x]}$, and the integral writes $\int_IF_{\delta_x}\mu(x)=F_\mu$. | |
| Mar 25 at 7:12 | comment | added | Pietro Majer | Can we say: for every finite Borel measure $\mu$ on $I:=[0,1]$ the integral $\int_Ifd\mu$ is the distribution function $t\mapsto \mu([t,1])$, so $\mu$ can be recovered from $\int_Ifd\mu$, and different measure give different integrals of $f$. | |
| Mar 24 at 18:43 | comment | added | Gerald Edgar | Not sure about that "countable image" assertion, in general. A nice example is $\sum_{k=1}^\infty (1/2)^k \epsilon_k$ where $\epsilon_k$ are independent random variables with $\epsilon_k = 1$ and $\epsilon_k = 0$ each with probability $1/2$. Then the sum is uniformly distributed on $[0,1]$. | |
| Mar 24 at 18:40 | vote | accept | Pietro Majer | ||
| Mar 24 at 18:36 | comment | added | Pietro Majer | In order to conclude, can’t we just say that $\sum_{k=1}^\infty\lambda_k \mathbf1_{[0,x_k]}$, seen as a true measurable function $[0,1]\to\mathbb R$, has countable image? On the contrary, any representative of $\psi\in L^2([01,])$ must have image of full measure in $[0,1]$ | |
| Mar 24 at 18:30 | history | edited | Gerald Edgar | CC BY-SA 4.0 |
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| Mar 24 at 18:23 | comment | added | Pietro Majer | Fantastic example! | |
| Mar 24 at 18:13 | history | edited | Gerald Edgar | CC BY-SA 4.0 |
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| Mar 24 at 18:07 | history | answered | Gerald Edgar | CC BY-SA 4.0 |