Need a function be continuous on a closed interval for the constitution of Intermediate Value Theorem? The statement for Intermediate Value Theorem is thus:

Let $f$ denote a function that is continuous on the closed interval $[a, b]$ and suppose $f(a)\neq f(b)$. If $N$ is any number between $f(a)$ and $f(b)$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

But who is to stop me if I put it like this?

Let $f$ denote a function that is continuous on the open interval $(a, b)$. Now, $\lim\limits_{x\to a^+} f(a)=L$ and $\lim\limits_{x\to b^-} f(b)=M$. Suppose, $L\neq M$. If $N$ is any number between $L$ and $M$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

 A: Nobody.
In fact, with $f$ as in your suggestion, define
$g\colon[a,b]\to \Bbb R$, $$g(x)=\begin{cases}f(x)&a<x<b\\L&x=a\\M&x=b\end{cases}$$
Then the classical IVT applied to $g$ is equivalent to your variant applied to $f$. With this in mind, the standard formulation looks perhaps more appealing.
A: Nothing. That's a slightly more general statement which is also correct.
A: Your version follows from the previous version. Let $F(x)=f(x)$ for $a <x<b$, $F(a)=L$ and $F(b)=M$. Then $F$ is continuous on $[a,b]$ and the first version applied to $F$ gives you the new version. 
A: The statement of intermediate value theorem is what you have writtem in the $1$st statement.But who told you that if the statement of intermediate value theorem is that one,then we cannot say that what you have written next is false.In fact if you take $I$ as any interval domain,then for any two points $x_1,x_2$ in $I$ consider restriction of $f$ on $[x_1,x_2]$,then apply intermediate value theorem,this will give you the generalized result.
i.e. If $f:I\to R$ is a continuous fucntion on an interval domain $I$ then if $f$ takes 2 values at $x_1$ and $x_2$,then it will take all values between the 2 values for some $x$ lying between $x_1$ and $x_2$.
