I'm reading this thread in which the OP asked if the following statement is true, i.e.,
Let $f$ be a non-negative measurable function and suppose that $f\in L^p(\Bbb R^n)$, then for any $\delta \gt 0$ there exists a non-negative continuous function with compact support, note as $h_\delta$, such that
(1) $\int |f(x)|^p dx – \delta^p \le \int|h_\delta(x)|^pdx$ and
(2) $h_\delta(x)\le f(x)$ for a.e. $x\in \Bbb R^n$.
Then @geetha290krm commented that
Take $f$ to the characteristic function of a fat Cantor set for a counter-example. In this case $h_\delta \equiv 0$.
I'm trying to formalize @geetha290krm's idea as follows.
Let $\lambda$ be the Lebesgue measure on $E := [0, 1]$. We work with $L_1 (E, \lambda, \mathbb R_{\ge 0})$. Let $C$ be the fat Cantor set. Then $\lambda(C)>0$. Let $f := 1_C$. Then $f=0$ on $E \setminus C$. Because $h_\delta (x) \le f (x)$ for $\lambda$-a.e. $x \in E$, there is a $\lambda$-null subset $N$ of $E$ such that $h_\delta \le f$ on $E \setminus N$. It follows that $h_\delta = 0$ on $E \setminus (C\cup N)$ and thus on $E \setminus (\overline C \cup N)$. Because $h_\delta$ is continuous, we get $h_\delta=0$ on $\overline{E \setminus (\overline C \cup N)}$. Because $C$ is nowhere dense, we get
$$
\overline{E \setminus \overline C} = E
\quad \text{and thus} \quad
\lambda \left ( \overline{E \setminus \overline C} \right ) = 1.
$$
To show that $h_\delta=0$ $\lambda$-a.e., it suffices to show that
$$
\lambda \left ( \overline{E \setminus (\overline C \cup N)} \right ) = 1.
$$
Then I got stuck. Could you elaborate how finish the proof?
Best Answer
Adding some details to my comment above:
If a continuous function is positive at a point it is positive in an open interval $I$ containing that point.
Thus, as noted in a comment by the OP, it suffices to show that an open interval $I$ is not contained in the union $C \cup N$, where $C$ is a fat Cantor set and $N$ is a Lebesgue null set.
Now $I \setminus C$ is nonempty and open, so it contains an open interval $J$, which cannot be covered by $N$.