Can I control Lebesgue integral by integration over small sets? I would like to know if for $f \in L^p(\Bbb R^n)$ for every $\epsilon >0$ there exists $\delta > 0$ such that for every measurable $E \subset \Bbb R^n$, $|E|<\delta$ it holds that $\int_{E} |f|^p~\mathrm{d}x < \epsilon$.
I also woud like to know if for each open set $\Omega \in \Bbb R^n$ and $f \in L^p(\Omega)$ it is possible to extend function f to $\Bbb R^n$ such that $F(x)=f(x) for x\in \Omega$ and $F(x)=0$ for $x \in \Bbb R^n \setminus \Omega$ and if always holds $F\in L^p(\Bbb R^n)$.
I suppose the answer for the last question is yes since sigma algebra of Lebesgue measurable sets contains all Borel sets and (thus also the set $\Bbb R^n \setminus \Omega$) and if  $\{x : f(x) > C \}$ is (Caratheodory) measurable on $\Omega$ it is also measurable on $\Bbb R^n$ since it is either $\{x : f_{\Omega}(x) > C \}$ or $\{x : f_{\Omega}(x) > C \} \cup \Bbb R^n \setminus \Omega$ but I do not know since in $\Bbb R^n$ there is more test sets than in $\Omega$.
 A: Let $f\in L^1$ and $f_n(x) = 1_{|f(x)|\le n}|f(x)|$, then $f_n\uparrow |f|$ and by monotone convergence one have that $\int f_n \uparrow \int |f|$. Now for given $\epsilon$, there exist an $n$ such that
$$
\int 1_{|f(x)|>n} |f(x)| {\rm d}x=\int |f(x)|{\rm d}x-\int f_n(x){\rm d}x \le \frac{\epsilon}{2}.
$$
Now let $\delta = \epsilon/2n$. Then if $m(A)\le \delta$, then
\begin{eqnarray}
\int_A |f(x)| {\rm d}x &= &\int_A 1_{|f(x)|>n} |f(x)| {\rm d}x+\int_A 1_{|f(x)|\le n}|f(x)| {\rm d}x
\\ &\le& 
\frac{\epsilon}{2} + m(A)\ n
\\ &\le& 
\frac{\epsilon}{2} + \frac{\epsilon}{2} = {\epsilon}.
\end{eqnarray}
For the case with $f\in L^p$, it is enough to substitute $|f|$ with $|f|^p$.
A: To answer the question in the first paragraph, yes: The following lemma is true:

If $g \in L^1(\mu)$, then for every $\epsilon > 0$, there exists $\delta > 0$ such that $$\mu(E) < \epsilon \implies \int_E |g| d\mu < \epsilon$$

This can be proved in quite a few ways; it's exercise $1.12$ in Rudin's Real and Complex Analysis, and it's proved as a theorem in Royden. Royden's proof has roughly the idea to approximate $g \ge 0$ by simple functions (which are bounded!), and then apply the Monotone Convergence Theorem. Then general $g$ can be handled by studying positive / negative or real / imaginary parts separately. Now your claim follows from the fact that $f \in L^p \iff f^p \in L^1$.
The answer to the second question is yes as well. Since $\Omega$ is open, the function
$$F(x) = \left\{\begin{array}{cc} f(x) &: x \in \Omega \\ 0 &: x \notin \Omega\end{array}\right.$$
is measurable (it can be viewed as the product $f \chi_{\Omega}$ of measurable functions), and $$\int_{\mathbb{R}^n} F d\mu = \int_{\Omega} F d\mu + \int_{\Omega^c} F d\mu = \int_{\Omega} f d\mu + \int_{\Omega^c} 0 \, d\mu = \int_{\Omega} f d\mu$$
and likewise for $p$-th powers.
A: Both of your statements are true. The second is very simple: products of measurable functions are measurable, continuous functions are measurable, and open sets are measurable. Your $F$ is just $f \chi_{\Omega}$.
The first is a little bit more involved but essentially follows from the definition of the Lebesgue integral, after some preliminary lemmas about approximation by simple functions. The proof:
Let $\varepsilon > 0$ and find a simple function of finite support $s$ with $\| f - s \|^p_p < \varepsilon/2$; we can assume the support is finite because $\mathbb{R}^n$ is $\sigma$-finite. (Royden/Fitzpatrick calls this the "Simple Approximation Theorem".) Then write $s=\sum_{k=1}^n a_k \chi_{A_k}$ and choose $\delta < \varepsilon/(2\max \{ a_1, \dots, a_n \})$. The rest is a short triangle inequality argument.
