Let $f:I \subset \mathbb{R}^n \rightarrow \bar{\mathbb{R}}$ be a Lebesgue measurable function and finite almost everywhere. If I is an interval, prove that given $\varepsilon$ and $\delta > 0$ there exists $\phi$ a simple function (finite linear combination of characteristic functions) such that
$ \vert \{x \in I: \vert \phi(x) - f(x)\vert > \varepsilon\} \vert \leq \delta$
I read this and assumed that $I$ is a compact interval. Then, redefining f to be zero where it was $\infty$, I can get a sequence of simple functions that converge pointwise to f. But, as $I$ is compact, the convergence is uniform. Then, the result is trivial.
However, I don't really know if I is a compact interval or if it goes to $\infty$ in any coordinate. In that case, is the result still valid? I can't get to prove it nor to find a counterexample.
Let $I=\mathbb{R}$ and $f(x)=x$. If $\phi$ is a simple function, then we can write $$ \phi=\sum_{i=1}^nc_i\chi_{E_i} $$ where each $c_i\in\mathbb{C}$ and the $E_i$ are disjoint measurable sets having finite measure.
In particular $\phi$ must be bounded, so there exists $B>0$ such that $|\phi(x)|\leq B$ for all $x\in\mathbb{R}$. By increasing $B$ if necessary, we can assume that $B>\varepsilon$, and it follows that if $|x|>2B$ then $$|f(x)-\phi(x)|\geq |x|-|\phi(x)|\geq 2B-B=B>\varepsilon$$
Therefore the set $\{x:|f(x)-\phi(x)|>\varepsilon\}$ contains the set $\{|x|>2B\}$, hence has infinite Lebesgue measure.
So the property you want certainly fails when $I$ is not a bounded interval. But even if $I$ is bounded, you might run into trouble if $f$ is unbounded on $I$.