The range of the Multiplication operator $R(T_f)=L^2(\mu)$ iff $\exists \epsilon > 0$ st. $|f|>\epsilon $ a.e. Let $(\Omega,\mu)$ be a sigma finite measure space and let $f:\Omega \rightarrow \mathbb{C}$ be a measurable function.
It is to show that the range of the Multiplication operator $R(T_f)=L^2(\mu)$ iff $\exists \epsilon > 0$ st. $|f|>\epsilon $ a.e.
I have already proven that the maximally defined multiplication operator is closed, its adjoint $ T_f^* = T_\bar{f} $ and that its kernel $\ker(T_f)=\{0\}$ iff $ \{ f=0 \} $ is a Nullset.
 A: Following my hint in the comments the "if" direction is trivial. It remains to show that if $R(T_f) = L^2(\mu)$ then there exists a $\varepsilon > 0$ such that $|f| > \varepsilon$ a.e.
Suppose not. Then for every $n \in \mathbb{N}$, $A_n = \{x: |f(x)|<n^{-1}\}$ has positive measure. It would be convenient to have the sets $A_n$ be disjoint and of positive, finite measures so next I show that it is possible to construct sets $C_n \subseteq A_n$ with those properties.
First, note that for every $n$, there is a $j > n$ such that $\mu(A_n \setminus A_j) > 0$. If not, $f 1_{A_n} = 0$ a.e. and then $1_{A_n} \not \in R(T_f)$. As a result of this, we can define a sequence $n_k$ by setting $$n_1 = 1, \qquad n_k = \min\{ j > n_{k-1}: \mu(A_{n_{k-1}} \setminus A_j) > 0\}$$
Then $B_j = A_{n_j} \setminus A_{n_{j+1}}$ is a set of positive measure such that $|f| < n_j^{-1} < j^{-1}$ on $B_j$. Further, since $A_{j+1} \subseteq A_j$, $B_j \cap B_k = \emptyset$ for $j \neq k$. The only reason that we cannot now set $C_j = B_j$ is that it may be that $\mu(B_j) = \infty$. However, since $(\Omega, \mu)$ is $\sigma$-finite, for each $j$ there is a $C_j \subseteq B_j$ such that $0 < \mu(C_j) < \infty$.
Having constructed these sets, we are ready to give an explicit $g \in L^2(\mu) \setminus R(T_f)$. Define $$g = \sum_{j=1}^\infty \frac{1}{j \mu(C_j)^\frac12} 1_{C_j}$$
Then $\|g\|_{L^2} = \sum_{j=1}^\infty \frac{1}{j^2} < \infty$ so that $g \in L^2(\mu)$. However, if $g \in R(T_f)$ then there is $h \in L^2(\mu)$ such that $g = hf$. Since $|f| < j^{-1}$ on $C_j$, we must then have that $|h| > \mu(C_j)^{-\frac12}$ on $C_j$. This means that $$\|h\|_{L^2} \geq \sum_{j=1}^\infty \int_{C_j} \mu(C_j)^{-1} d\mu = \infty$$ which is a contradiction.
