I would like to prove that the union of two subspaces is equal to a subspace.
note $X$ a measurable space and $f,g,h : X \rightarrow \mathbb{R}$ are mesurable functions with also $r ,t >0$
I would like to prove that $E_1 = \{x \in X | |f-g| \geq r \} $ and $ E_2 = \{ x \in X | |g-h| \geq t \}$ gives $E_1 \cup E_2 = \{ x \in X | |f-h| \geq r +t \}$
I don't know as to how I should proceed to prove the following
If $x \notin E_1$ and $x \notin E_2$, then $|f(x) - h(x)| \le |f(x) - g(x)| + |g(x) + h(x)| < r + t$, so $x \notin \{y \in X \mid |f(y) - h(y) \ge r+t\}$. This proves $$E_1^c \cap E_2^c \subseteq \{x \in X \mid |f(x) - h(x) \ge r+t\}^c$$ which is equivalent to $$E_1 \cup E_2 \supseteq \{x \in X \mid |f(x) - h(x) \ge r+t\}.$$
I don't think the reverse inclusion holds. Let $f \equiv 1$, $g \equiv 0$, and $h \equiv 1$, and let $r=t=1/2$. Then $E_1=E_2 = X$, but $\{x \in X \mid |f(x) - h(x)| \ge 1\} = \varnothing$.