Show that there is no continuous function f s.t $f=g$ a.e when $g=1_{[0,\infty)}$ So I have to show that there can be no continuous function f s.t $f=g$ a.e when $$g=1_{[0,\infty)}$$
All i know is the following:
$g=0$ for $x<0$ and $g=1$ for $$0\leq x<\infty$$
If f is continuous at f(0) then there must exist a mapping from $(-\delta,\delta)$ to $(f-\epsilon,f + \epsilon)$ for any $\delta,\epsilon > 0$.
I have no idea how to show that this cannot be the case..
 A: Assume such an $f$ exists. Then, clearly, $\lim_{x\to-\infty} f(x)=0$ and $\lim_{x\to\infty}f(x)=1$, so there must be some real number $c$ such that $f(c)=1/2$, by the intermediate value theorem.
By definition of continuity and $\epsilon=1/4$, there must be a $\delta>0$ such that on $(c-\delta, c+\delta)$, $f$ takes only values between $1/4$ and $3/4$. In other words, for any $x\in(c-\delta, c+\delta)$, we have $f(x)\neq g(x)$. Since that interval has measure $2\delta$, we cannot have $f=g$ a.e.
A: Suppose it is. Then, $m\{x\mid f(x)\notin\{0,1\}\} =0$. By continuity of $f$, $$f^{-1}(]-\infty ,0[\cup]0,1[\cup]1,\infty [)$$
is open (and of measure $0$). Therefore  $$f^{-1}(]-\infty ,1[\cup]1,0[\cup]1,\infty [)=\emptyset$$
and thus $f(x)=g(x)$ for all $x$ what contradict the continuity of $f$.
A: Let's prove it by contradiction: assume f exists.
We will get our focus on point 0.
if f exists then because f = g almost everywhere that means that however close you won't come from the minus side, you'll be able to find a point x where f(x) = g(x) = 0 (otherwise there would be an open (-e, 0) on which f is different then g, which is contradiction), so you create a sequence from these points coming to zero
with the same argument but from the right side, you show that there exists a sequence of points that come to zero from the right side (on every each of them f is evaluated as 1.
therefore these are two sequences in which f have different limits and they come to the same point (which is 0). so f cannot be continuous, and that is our contradiction.
A: Let $f:\mathbb R\to\mathbb R$ denote a continuous function.
Automatically $f(0)\neq1\vee f(0)\neq0$.
If $f(0)\neq1$ then $f(x)\neq1$ on some interval $(-\epsilon,\epsilon)$ so that $f(x)\neq g(x)$ for $x\in[0,\epsilon)$. 
If $f(0)\neq0$ then $f(x)\neq0$ on some interval $(-\epsilon,\epsilon)$ so that $f(x)\neq g(x)$ for $x\in(-\epsilon,0)$. 
(In the statements above $\epsilon$ is positive)
In both cases $\lambda(\{f\neq g\})\geq\epsilon>0$, hence $f=g$ a.e. is not true.
