Let $\langle X,\mu\rangle$ be a measure space and for $f\colon X\rightarrow \mathbb{C}$ define $$ \operatorname{Ess\, Im}(f):=\left\{ y\in \mathbb{C}:\mu (f^{-1}(B_{\varepsilon}(y)))>0\text{ for all }\varepsilon >0\text{.}\right\} . $$ (This is the essential image of $f$, also known as the essential range.)

Let $f,g\colon X\rightarrow \mathbb{C}$. As $\operatorname{Im}(f+g)\subseteq \operatorname{Im}(f)+\operatorname{Im}(g)$, I conjectured that also $\operatorname{Ess\, Im}(f+g)\subseteq \operatorname{Ess\, Im}(f)+\operatorname{Ess\, Im}(g)$, but I have found this quite a bit more difficult to prove than I imagined. How might I go about doing this?


First let me, for ease of notation, denote the essential image of a function $f$ by $\mathcal{R}(f)$.
The statement is wrong in its full generality (see a counterexample below). However it is true when $\mathcal{R}(f)+\mathcal{R}(g)$ is a closed subset of $\mathbb{C}$. Note in particular that this condition is satisfied when $f$ and $g$ are bounded.

Take any $z \notin \mathcal{R}(f)+\mathcal{R}(g)$, by closedness, $$ \quad \quad \quad \quad \exists \epsilon >0: B(z,\epsilon) \subset \left( \mathcal{R}(f)+\mathcal{R}(g) \right)^c \quad \quad \quad \quad (1)$$ (here $^c$ denotes the complement). Take any $x \in X$ such that $(f +g)(x) \in B(z, \epsilon)$. By $(1)$ we must have that either $f(x) \notin \mathcal{R}(f)$ or $g(x) \notin \mathcal{R}(g)$. So we get that $$ (f+g)^{-1}(B(z,\epsilon)) \subset f^{-1}(\mathcal{R}(f))^c \cup g^{-1}(\mathcal{R}(g))^c. $$ From this we have that $$ \mu \left( (f+g)^{-1}(B(z,\epsilon)) \right) \leq \mu \left( f^{-1}(\mathcal{R}(f))^c\right) +\mu \left( g^{-1}(\mathcal{R}(g))^c \right). $$ Now observe that for any function $f: \mu \left( f^{-1}(\mathcal{R}(f))^c\right) = 0$ (in fact $f(A) \cap \mathcal{R}(f) = \emptyset \Rightarrow \mu(A)=0$), hence $z$ can't be an element of the essential range of $f+g$. So $\mathcal{R}(f+g) \subset \mathcal{R}(f) + \mathcal{R}(g)$.

Let $\varphi: \mathbb{N} \rightarrow \mathbb{Q}$ be any surjection and note that if I write $\varphi(n) = \frac{p}{q}$. I always take the fraction so that $p$ and $q$ are coprime and $p$ is positive. Extend $\varphi$ on $\mathbb{Z}$ by letting $\varphi(-n) = \varphi(n)$, but with the convention that $p$ is negative. Define two maps: $$f: \mathbb{R} \rightarrow \mathbb{R}: x \mapsto p, \text{ where } p \text{ is the numerator of } \varphi(n) \text{ where } n \text{ satisfies } n \leq x < n+1, $$ $$g: \mathbb{R} \rightarrow \mathbb{R}: x \mapsto q \sqrt{2}, \text{ where } q \text{ is the denominator of } \varphi(n) \text{ where } n \text{ satisfies } n \leq x < n+1. $$ Then $f+g$ is the map $x \mapsto p+ q\sqrt{2}$.
By Dirichlet's aprroximation Theorem $\{p+ q\sqrt{2} \mid p, q \in \mathbb{Z} \text{ coprime} \}$ is dense in $\mathbb{R}$. Then if $z \in \mathbb{R}$ and $\epsilon>0$, there exists $p,q \in \mathbb{Z}$ coprime such that $|p+ q \sqrt{2} - z| < \epsilon$. Let $n \in \mathbb{Z}$ be such that $\varphi(n) = \frac{p}{q}$. Then $[n,n+1) \subset (f+g)^{-1}(B(z,\epsilon))$, so that $\mathcal{R}(f+g) = \mathbb{R}$. However $\mathcal{R}(f) + \mathcal{R}(g) = \{ p +q \sqrt{2} \mid p,q \in \mathbb{Z} \text{ coprime} \}$ which does not contain $\mathbb{R}$.


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