Prove that If f is continuous function, then Range of f = Essential range of f
I start by : Assume that y $\in$ R(f),then show that
M($f^{-1}$ (y, €)) $\gt$ 0
(y, €) is open interval, then since f is continuous
$f^{-1}$ (y, €) is open interval also. So M($f^{-1}$ (y, €)) = length of (y, €) $\gt$0
This leads to R(f) $\subseteq$ of Ess_R(f)
Now, how can I prove the opposite?
You have not specified the domain of $f$. The function $e^{-|x|}$ is continuous on $\mathbb R$. $0$ is in its essential range but not in its range. However, if $f$ is defined an a compact interval $[a,b]$ then its range coincides with its essential range: suppose $y$ belongs to the essential range of $f$. For each $n$ there exists a point $x_n$ in $f^{-1}(y-\frac 1n , y+\frac 1 n)$. So $|f(x_n)-y| <\frac 1 n$ for each $n$. There is a subsequence of $(x_n)$ which converegs to some point $x$ and continuity of $f$ gives $y =f(x)$.