Equality on the Closure

Detailed Solution

Define $$ h(x)=f(x)-g(x). $$ Since $f$ and $g$ are continuous, $h$ is continuous on $\mathbb R$. The given condition $$ f(x)=g(x)\qquad\text{for all }x\in U $$ means that $$ h(x)=0\qquad\text{for all }x\in U. $$ We now consider any point $a\in\overline{U}$. By the definition of closure, there exists a sequence $(x_n)$ with $$ x_n\in U $$ for every $n$ and $$ x_n\to a. $$ Since $h$ is continuous, $$ h(a)=\lim_{n\to\infty}h(x_n). $$ But $x_n\in U$, so $$ h(x_n)=0 $$ for every $n$. Hence $$ h(a)=\lim_{n\to\infty}0=0. $$ Therefore, $$ f(a)=g(a) $$ for every $a\in\overline{U}$. Thus, $$ \boxed{f\equiv g\text{ on }\overline{U}}. $$ Equality outside $\overline{U}$ is not forced. For example, take $$ U=(0,1), $$ and define $$ f(x)=0,\qquad g(x)= \begin{cases} 0,&0\le x\le1,\\ (x-1)^2,&x>1. \end{cases} $$ Both functions are continuous, and they agree on $\overline U=[0,1]$, but they differ for $x>1$. Now consider the remaining options. A is false because equality on a non-empty open set does not imply equality on all of $\mathbb R$. C is false because continuity does provide equality at every boundary point of $U$, as shown above. D is false because an interval containing $U$ may extend beyond $\overline U$, where $f$ and $g$ need not agree. Therefore, the only statement that must be true is $$ \boxed{\mathrm{B}}. $$