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Suppose that $S$ is the graph of a continuous, strictly increasing function $\psi$. Then $S(W) = \psi(W)$, and the question asks if there is $\psi$ and $W$ such that $|W| = 1$, but $|\psi(W)| = 0$. The answer is affirmative.

Let $\mu$ be a singular probability measure on $(0, 1)$ with no atoms and full support, and let $\phi(x) = \mu((0, x))$ denote its distribution function, and $\psi = \phi^{-1}$ be the inverse function. Then $\mu(A) = |\phi^{-1}(A)| = |\psi(A)|$ for every $A = (0, x)$, and hence for every Borel $A$. Let $V$ be a Borel set such that $|V| = 0$ and $\mu(V) = 1$. Then $W = (0, 1) \setminus V$ satisfies $|W| = 1$ and $|\psi(W)| = \mu(W) = 0$, as desired.

Suppose that $S$ is the graph of a continuous, strictly increasing function $\psi$. Then $S(W) = \psi(W)$, and the question asks if there is $\psi$ and $W$ such that $|W| = 1$, but $|\psi(W)| = 0$. The answer is affirmative.

Let $\mu$ be a singular probability measure on $(0, 1)$ with no atoms and full support, and let $\psi(x) = \mu((0, x))$ denote its distribution function. Then $\mu(A) = |\psi(A)|$ for every $A = (0, x)$, and hence for every Borel $A$. Let $V$ be a Borel set such that $|V| = 0$ and $\mu(V) = 1$. Then $W = (0, 1) \setminus V$ satisfies $|W| = 1$ and $|\psi(W)| = \mu(W) = 0$, as desired.

Suppose that $S$ is the graph of a continuous, strictly increasing function $\phi$. Then $S(W) = \phi(W)$, and the question asks if there is $\phi$ and $W$ such that $|W| = 1$, but $|\psi(W)| = 0$. The answer is affirmative.

Let $\mu$ be a singular probability measure on $(0, 1)$ with no atoms and full support, and let $\phi(x) = \mu((0, x))$ denote its distribution function, and $\psi = \phi^{-1}$ be the inverse function. Then $\mu(A) = |\phi^{-1}(A)| = |\psi(A)|$ for every $A = (0, x)$, and hence for every Borel $A$. Let $V$ be a Borel set such that $|V| = 0$ and $\mu(V) = 1$. Then $W = (0, 1) \setminus V$ satisfies $|W| = 1$ and $|\psi(W)| = \mu(W) = 0$, as desired.

Suppose that $S$ is the graph of a continuous, strictly increasing function $\psi$. Then $S(W) = \psi(W)$, and the question asks if there is $\psi$ and $W$ such that $|W| = 1$, but $|\psi(W)| = 0$. The answer is affirmative.

Let $\mu$ be a singular probability measure on $(0, 1)$ with no atoms and full support, and let $\phi(x) = \mu((0, x))$ denote its distribution function, and $\psi = \phi^{-1}$ be the inverse function. Then $\mu(A) = |\phi^{-1}(A)| = |\psi(A)|$ for every $A = (0, x)$, and hence for every Borel $A$. Let $V$ be a Borel set such that $|V| = 0$ and $\mu(V) = 1$. Then $W = (0, 1) \setminus V$ satisfies $|W| = 1$ and $|\psi(W)| = \mu(W) = 0$, as desired.

Suppose that $S$ is the graph of a continuous, strictly increasing function $\phi$. Then $S(W) = \phi(W)$, and the question asks if there is $\phi$ and $W$ such that $|W| = 1$, but $|\phi(W)| = 0$. The answer is affirmative.

Let $\mu$ be a singular probability measure on $(0, 1)$ with no atoms and full support, and let $\phi(x) = \mu((0, x))$ denote its distribution function. Then $\mu(A) = |\phi(A)|$ for every $A = (0, x)$, and hence for every Borel $A$. Let $V$ be a Borel set such that $|V| = 0$ and $\mu(V) = 1$. Then $W = (0, 1) \setminus V$ satisfies $|W| = 1$ and $|\phi(W)| = \mu(W) = 0$, as desired.

Suppose that $S$ is the graph of a continuous, strictly increasing function $\phi$. Then $S(W) = \phi(W)$, and the question asks if there is $\phi$ and $W$ such that $|W| = 1$, but $|\psi(W)| = 0$. The answer is affirmative.

Let $\mu$ be a singular probability measure on $(0, 1)$ with no atoms and full support, and let $\phi(x) = \mu((0, x))$ denote its distribution function, and $\psi = \phi^{-1}$ be the inverse function. Then $\mu(A) = |\phi^{-1}(A)| = |\psi(A)|$ for every $A = (0, x)$, and hence for every Borel $A$. Let $V$ be a Borel set such that $|V| = 0$ and $\mu(V) = 1$. Then $W = (0, 1) \setminus V$ satisfies $|W| = 1$ and $|\psi(W)| = \mu(W) = 0$, as desired.