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Let $F$ be a CDF of random variable $X$ distributed in $[0,1]$. $F$ has a density, $f$. Let $\mu_F$ denote the corresponding measure, i.e. $\mu_F(A)=\int\limits_{A}f(x)dx$ for all $X \in \mathcal{B}([0,1])$, the Borel-sigma algebra on $[0,1]$.

Let $g:[0,1] \rightarrow \mathbb{R}$ be a continuous function with no flat regions.

I want to show that $\mu_F(\{g(x)\leq y\})$ is continuous in $y$.

In the case where $g(x)=y$ has finite no. of solutions for all $y \in [\min\limits_{x\in[0,1]}g(x),\max\limits_{x\in[0,1]}g(x)]$, I can show this. But I'm not able to manage the case where it has infinite no. of solutions. I asked this question as a step forward. But I'm still not able to solve that case. Any help is most appreciated.

Canine360
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1 Answers1

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That's not true. Let $T$ be closed positive measure nowhere dense subset of $[0, 1]$ (for example, fat Cantor set) and define $g(x) = \rho(x, T)$.

To prove that $g$ is continuous let's show that if $|x - y| < \epsilon / 2$ then $|g(x) - g(y)| < \epsilon$.

Let $s = g(x)$. By definition of $g$, there is $z \in T$ s.t. $|x - z| < s + \epsilon / 2$. Then $g(y) \leq |z - y| \leq |z - x| + |x - y| < s + \epsilon$.

If $g(y) < s - 3 \epsilon / 4$ then there is $v \in T$ s.t. $|y - v| < s - \epsilon / 2$ and then $|x - v| \leq |x - y| + |y - v| < \epsilon / 2 + s - \epsilon / 2 < s$. But $g(x) = s$, so there is no point in $T$ with distance to $x$ less than $s$ - contradiction, so $g(y) > s - 3 \epsilon / 4$.

So, $g(x) - 3 \epsilon / 4 < g(y) < g(x) + \epsilon$, thus $|g(y) - g(x)| < \epsilon$.

To prove that $g$ has no flat region, consider any point $x$ and let's show that $g$ isn't constant near $x$.

If $x \in T$ then this is simple: $g(x) = 0$, but as $T$ is nowhere dense, there is $y$ near $x$ s.t. $y \notin T$. As $T$ is closed this implies $g(y) > 0$.

If $x$ is left of all points of $T$, then moving a bit left increases distance to $T$, so $g$ isn't constant near $x$, similarly if $x$ is right to $T$.

Finally, if there are points of $T$ both on left and right of $x$, let $a$ be $\sup T \cap [0, x]$ and $b$ be $\inf T \cap [x, 1]$ (ie closest points to $x$ from left and right) - such points exist because $T$ is closed and bounded, thus compact. If $|x - a| \leq |x - b|$ then going a bit left from $x$ decreases $g$, and if $|x - a| > |x - b|$ then going a bit right from $x$ decreases $g$, so again $g$ isn't constant near $x$.

So, $g$ is continuous and has no flat regions. Also, note that $g$ maps $T$ to zero.

Let $\mu_F$ be Lebesgue measure (corresponds to uniform distribution, which is continuous), then we have $\mu_F(\{g(x) \leq -\varepsilon\}) = 0$ for any $\varepsilon > 0$, but $\mu_F(\{g(x) \leq 0\}) = \mu(T) > 0$.

mihaild
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