Approximate a bounded function from below by step functions I'm curious in the following problem:
Let $f:[0,1]\to \mathbf R_+$ be a bounded and measurable function. It is well known that we can approximate $f$ with simple functions in a "monotone" way; that is, we can find a sequence of simple functions $\{\phi_k\}$ such that
$1$. $\phi_{n}\le\phi_{n+1}$ for all $n\in\mathbf N$ on $[0,1]$; and
$2$. $\phi_k(x)\to f(x)$ for each $x\in[0,1]$ as $k\to\infty$.
However, we are not able to do the same thing with a sequence of step functions (a step function is a simple function such that each indicator function it involves in its representation is supported by an interval). The best result I know is that we can find a sequence of step functions that converges pointwise to $f$ almost everywhere.
Thus, I was wondering if we can strengthen the case with step functions a bit. In particular, can we always find a sequence of step functions $\{\psi_k\}$ which converges to $f$ pointwise almost everywhere on $[0,1]$, and satisfies
$f(x)\ge \psi_k(x)$ a.e. on $[0,1]$ for each $k$?
Any hint/advice would be much appreciated.
 A: The answer is no and here is a counter-example.
Let $C$ be any closed subset of $[0, 1]\setminus \mathbb Q$  with positive measure.  For example
$$
  C=[0,1]\setminus \bigcup_{n\in {\mathbb N}}\big (q_n-\frac\varepsilon {2^n},q_n+\frac\varepsilon {2^n}\big ),
  $$
where $\varepsilon $ is very small and $\{q_n\}_{n\in {\mathbb N}}$ is an enumeration of the rational numbers in $[0,1]$.
We then claim that the characteristic
function $1_C$ is a counter-example to the conjecture by the OP.
In order to prove it, it is enough to show that there is no nonnegative simple function $\psi $
satisfying $\psi (x)\leq 1_C(x)$, for almost every $x$, unless of course $\psi $ is a linear combination of degenerate intervals of the form
$[a,a]$.
Arguing by contradiction, suppose otherwise.  So, in particular, there is a non-degenerate interval $(a,b)\subseteq [0,
1]$, and some $\lambda >0$ such that
$$
  \lambda 1_{(a, b)}(x)\leq  1_C(x), \tag{$\dagger$}
  $$
for almost every $x$.  This means that there is some null set $Z\subseteq [0,1]$, such that for every $x$ in $[0, 1]\setminus
Z$, one has that ($\dagger$) holds.  In particular, if $x\in (a,b)\setminus Z$, one has that $0<\lambda \leq  1_C(x)$, so $x\in  C$.
In other words,
$$
  (a,b)\setminus Z\subseteq  C.
  $$
Since $Z$ has measure zero, its interior is empty, so   $(a,b)\setminus Z$ is  dense in $(a,b)$ and  it then follows that
$$
  (a,b)\subseteq \overline{(a,b)\setminus Z} \subseteq C,
  $$
but this contradicts the fact that $C$ contains no rational numbers.
