If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$? Suppose $f\in L^1(\mathbb{R})\cap L^\infty(\mathbb{R})$ is bounded, non-negative and integrable (w.r.t. Lebesgue measure) : does there exist $g$ continuous (non-negative) and integrable such that
$$ f\leq g$$
almost everywhere? If so, can we choose $g$ bounded on top of that?
 A: In general, there does not exist such a $g$. To see this, let $(q_n)_{n \in \Bbb{N}}$ be an enumeration of $\Bbb{Q}$. Using $\sigma$-subadditivity of the Lebesgue measure, it is easy to see that the open set
$$
U := \bigcup_{n \in \Bbb{N}} (q_n - 2^{-n}, q_n + 2^{-n})
$$
has finite measure; in fact, $\lambda(U) \leq 2$. Hence, the indicator function/characteristic function $f := 1_U$ satisfies all of your assumptions.
Now, assume that $f \leq g$ almost everywhere with $g$ continuous, say $f(x) \leq g(x)$ for all $x \in \Bbb{R} \setminus N$, with $\lambda(N) = 0$.
We will show $g \geq 1$, so that $g$ is not integrable. By density of $\Bbb{Q}$, it suffices to show $g(x) \geq 1$ for all $x \in \Bbb{Q}$. Let $x \in \Bbb{Q}$ be arbitrary. It suffices to show that for each $\varepsilon > 0$, there is $x_\epsilon$ with $|x - x_\varepsilon| < \varepsilon$ and $g(x_\varepsilon) \geq 1$.
But $x = x_n$ for some $n \in \Bbb{N}$ and the interval
$$
I := (x - \varepsilon, x + \varepsilon) \cap (x - 2^{-n}, x + 2^{-n})
$$
has positive measure, so that there is some $y \in I \setminus N$. But $I \subset U$, so that $1 = f(y) \leq g(y)$. All in all, $x_\varepsilon := y$ does what we want.
A: In general, there does not exists such a $g$. For example, let $f\colon \mathbf R \to \mathbf R$ defined by 
$$ f(x) = \frac 1{1+x^2} + \chi_{\mathbf Q}(x) $$
Then $f \in L^1(\mathbf R) \cap L^\infty(\mathbf R)$ is non-negative, but any $g \in C^0(\mathbf R)$ with $g \ge f$ also fulfills $g \ge \chi_{\mathbf Q}$ and hence $g \ge 1$ due to continuity, and hence $g \not\in L^1(\mathbf R)$.
A: Yes, I think so. Here's an idea of a proof, but there are some issues that I've not resolved:
Let $E_n = \{x : n-1 < f(x)<n\}$ which are all measurable with finite measure, we also have that for large enough $E_n$ that $\mu(E_n) = 0$. We also have that $f \le \sum \chi_j$ a.e. where $\chi_j$ is the characteristic function of $E_j$.
So we have to prove that there's a continuous function $g_j \ge \chi_j$ that is integrable. We want to do that by setting $g_j = 1$ on $E_j$ and that it's support is measurable with limited measure (ie less than $\mu(E_j)+\epsilon_j$ for an $\epsilon_j$ of our choice). Then we would have that $\int g dx < \sum \int \chi_j + \sum \epsilon_j < \infty$  if we select $\epsilon_j$ wisely.
Now since $E_j$ is measurable we have that there's an open set $U_j\supset E_j$ such that $\mu(U_j)<\mu(E_j)+\epsilon_j/2$. Now we use that we can write $U_j$ as an union of contable non-intersecting (open) intervals, $U_j = \bigcup I_j(k)$.
Now we form functions that are $1$ on $I_j(k)$ and drops of fast enough so it's integral is no larger than $\mu(I_j(k))+\epsilon 2^{-k}$. This construct should hopefully be enough to make sure that their sum is continuous and it's integral would be small enough.
Then adding all these functions up would give us $g$.
