<p>Let $n=1$. Let $\mu$ be the usual Lebesgue length measure on $[1/2,1]$, and let $\mu$ be the <strong><em>negative</em></strong> of the usual Lebesgue length measure on $[0,1/2]$.</p>

<p>In particular, Lebesgue measure is $|\mu|$.</p>

<p>Let $\mathcal U\subseteq [0,1]$ be a set of Lebesgue measure 0. (For instance, $\mathcal U$ could be the set of all numbers whose binary expansion does not satisfy the law of large numbers...)</p>

<p>Let</p>

<ul>
<li>$\sigma(x)=1$ for $x\in\mathcal U$, and</li>
<li>$\sigma(x)=0$ for $x\not\in\mathcal U$.</li>
</ul>

<p>Then given $y$ and $\theta$, there are two cases.</p>

<p><strong>Case 1</strong>: $y\ne 0$. Then
$$|\mu|\{X:y^\top X+\theta\in\mathcal U\}=0,$$ hence
$$|\mu|\{X:\sigma(y^\top X+\theta)\ne 0\}=0,$$
so $\int \sigma(y^\top X+\theta)d|\mu|(x)=0$,
so $\int \sigma(y^\top X+\theta)d\mu(x)=0$.</p>

<p><strong>Case 2</strong>: $y=0$. Then $\sigma(y^\top X+\theta)$ is the constant $\sigma(\theta)$, which satisfies
$$\int \sigma(\theta)\,d\mu(x)=\sigma(\theta)\int d\mu(x)=\sigma(\theta)\mu([0,1])=0.$$</p>

<p>So, $\sigma$ is non-discriminatory.</p>

CC BY-SA 4.0
2018-08-22T07:08:55.260Z
2018-08-22T07:08:55.260Z
Bjørn Kjos-Hanssen
<p>I am reading Cybenko's "Approximation by Superpositions of a Sigmoidal Function".
The paper defines a discriminatory function as:</p>

<p>$\sigma$ is <strong><em>discriminatory</em></strong> if for a measure $\mu$,
\begin{align}
\int \sigma (y^\top x + \theta) d \mu(x) = 0
\end{align}
for all $y \in R^n$ and $\theta \in R$, then $\mu=0$.</p>

<p>What is an example (with proof) of a non-discriminatory function? (besides 0)</p>

CC BY-SA 4.0
2018-08-22T02:29:35.787Z
What about $\sigma$=the sign function? Then for $\mu$ which is a product of normal gaussians, the integral is zero.
Aryeh
CC BY-SA 4.0
2018-08-22T03:08:58.237Z
well in that case if we let $\theta$ go to infinity, then wouldnt the output of $\sigma$ be 1 for most of the mass, so the integral would be positive?
LYH
CC BY-SA 4.0
2018-08-22T03:11:05.613Z
Can you spell out the exact quantifiers over $y$, $\theta$, and $\mu$?
Aryeh
CC BY-SA 4.0
2018-08-22T03:25:05.830Z
$\mu$ is some finite signed regular Borel measure on $[0, 1]^n$. (you only have to find one $\mu$ that works)
The condition has to be satisfied for any $y \in R^n$ and any $\theta \in R$.

I'm confused how a nontrivial function can make that integral $0$ for all $y$ and all $\theta$.
LYH
CC BY-SA 4.0
2018-08-22T04:06:19.613Z
I agree, in this formulation it doesn't seem like there should be any non-trivial candidates for $\mu$. I suspect it's an issue of quantifiers.
Aryeh
<machine-learning><approximation><function>
Examples of nontrivial non-discriminatory functions
LYH
# Examples of nontrivial non-discriminatory functions

(Asked by: username_0 on 2018-08-22)

I am reading Cybenko's "Approximation by Superpositions of a Sigmoidal Function".
The paper defines a discriminatory function as:

$\sigma$ is ***discriminatory*** if for a measure $\mu$,
\begin{align}
\int \sigma (y^\top x + \theta) d \mu(x) = 0
\end{align}
for all $y \in R^n$ and $\theta \in R$, then $\mu=0$.

What is an example (with proof) of a non-discriminatory function? (besides 0)

**Tags:** machine-learning, approximation, function
> **Comments:**
> **username_1**: What about $\sigma$=the sign function? Then for $\mu$ which is a product of normal gaussians, the integral is zero.  
> **username_0**: well in that case if we let $\theta$ go to infinity, then wouldnt the output of $\sigma$ be 1 for most of the mass, so the integral would be positive?  
> **username_1**: Can you spell out the exact quantifiers over $y$, $\theta$, and $\mu$?  
> **username_0**: $\mu$ is some finite signed regular Borel measure on $[0, 1]^n$. (you only have to find one $\mu$ that works) The condition has to be satisfied for any $y \in R^n$ and any $\theta \in R$.  I'm confused how a nontrivial function can make that integral $0$ for all $y$ and all $\theta$.  
> **username_1**: I agree, in this formulation it doesn't seem like there should be any non-trivial candidates for $\mu$. I suspect it's an issue of quantifiers.  

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## 1 answers

### Answer by username_2 on 2018-08-22. Score: 1 | Accepted answer

Let $n=1$. Let $\mu$ be the usual Lebesgue length measure on $[1/2,1]$, and let $\mu$ be the ***negative*** of the usual Lebesgue length measure on $[0,1/2]$.

In particular, Lebesgue measure is $|\mu|$.

Let $\mathcal U\subseteq [0,1]$ be a set of Lebesgue measure 0. (For instance, $\mathcal U$ could be the set of all numbers whose binary expansion does not satisfy the law of large numbers...)

Let

* $\sigma(x)=1$ for $x\in\mathcal U$, and
* $\sigma(x)=0$ for $x\not\in\mathcal U$.

Then given $y$ and $\theta$, there are two cases.

**Case 1**: $y\ne 0$. Then
$$|\mu|\{X:y^\top X+\theta\in\mathcal U\}=0,$$ hence
$$|\mu|\{X:\sigma(y^\top X+\theta)\ne 0\}=0,$$
so $\int \sigma(y^\top X+\theta)d|\mu|(x)=0$,
so $\int \sigma(y^\top X+\theta)d\mu(x)=0$.

**Case 2**: $y=0$. Then $\sigma(y^\top X+\theta)$ is the constant $\sigma(\theta)$, which satisfies
$$\int \sigma(\theta)\,d\mu(x)=\sigma(\theta)\int d\mu(x)=\sigma(\theta)\mu([0,1])=0.$$

So, $\sigma$ is non-discriminatory.


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