# Essential image of sum

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 $$\mathbb{R}$$.