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Show that if f and g are monotone non-increasing real-valued functions on an open real interval I which agree on a dense subset of I, and g is continuous, then f = g on I,

and give an example to show the conclusion is not valid if the assumption that g is continuous is omitted.

Attempt: Is it similar to this:$f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$ ?

Tim
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  • It is similar, but you need to take advantage of I being an ordered set. Every number falls between sequences approaching that number, as well as the non-increasing part. – Dunham Sep 29 '19 at 18:14

1 Answers1

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Let $D=\{x\in I: f(x)=g(x)\}$. Take any $a\in I$. Take increasing $c_n\in D$ and decreasing $d_n\in D$ with $c_n\to a$ and $d_n\to a$. By continuity, $g(c_n)\to g(a)$ and $g(d_n)\to g(a)$. Since $f$ is non-increasing, we have $g(c_n)=f(c_n)\ge f(a)$ and $f(a)\ge f(d_n)=g(d_n)$ for all $n$. Taking limits we obtain $g(a)\ge f(a)\ge g(a)$, thus $f(a)=g(a)$.

Re an example, take
$f(x)= \begin{cases} 1 \ \mathrm{if}\ 0< x\le \frac12 \\ 0 \ \mathrm{if}\ \frac12<x<1\end{cases}$ and
$g(x)= \begin{cases} 1 \ \mathrm{if}\ 0<x<\frac12 \\ 0 \ \mathrm{if}\ \frac12\le x<1\end{cases}$.
Note that $f$ and $g$ agree, and are continuous on the open interval $I=(0,1)$, except at $x=\frac12$, where both functions are discontinuous and $f(\frac12)\neq g(\frac12)$.

Mirko
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