If a function is Riemann integrable, then it is Lebesgue integrable and 2 integrals are the same? Is is true that if a function is Riemann integrable, then it is Lebesgue integrable with the same value? If it's true, how to prove it? If it's false, what is a counterexample?
 A: If you are talking about proper Riemann integrals, i.e. $f : [a,b] \to \Bbb{R}$ is bounded and the interval $[a,b]$ is compact, then this is true.
EDIT: In the following, all integrals $\int \dots \, dx$ are to be unterstood as the Riemann integral $\int_a^b \dots \, dx$. All integrals $\int \dots \, d\lambda(x)$ are to be understood as the Lebesgue integral $\int_{[a,b]} \dots \, \lambda(x)$.
For a proof, use that there are sequences $(\varphi_n)_n$ and $(\psi_n)_n$ of Riemann step functions such that $\varphi_n \leq f \leq \psi_n$ and
$$\int \varphi_n \, dx \to \int f \, dx \leftarrow \int \psi_n \, dx.$$
Depending on your exact definition of the Riemann integral, this is either a direct consequence, or an easy consequence of the definition.
By changing to $\max\{ \varphi_1, \dots, \varphi_n \}$ and $\min\{\psi_1, \dots, \psi_n\}$, we can assume w.l.o.g. that the sequences  $(\varphi_n)_n$ and $(\psi_n)_n$ are increasing/decreasing.
On Riemann step functions $\gamma : [a,b] \to \Bbb{R}$, the Lebesgue integral and the Riemann integral coincide (why?). Hence,
$$
\int |\psi_n - \varphi_n| \, d\lambda(x) = \int \psi_n - \varphi_n \, dx \to \int f\, dx - \int f \, dx = 0. \qquad (\dagger)
$$
By monotonicity, also $\varphi_n \to \varphi$ and $\psi_n \to \psi$ pointwise with $\varphi_1 \leq \varphi \leq f \leq \psi \leq \psi_1$.
By dominated convergence,
$$
\int |\psi_n - \varphi_n| \, d\lambda(x) \to \int |\psi - \varphi| \, d\lambda(x).
$$
By $(\dagger)$, we get $\int|\psi- \varphi|\, d\lambda(x) = 0$ and hence $\psi = \varphi$ almost everywhere.
Because of $\varphi \leq f \leq \psi$, we get $f = \varphi = \psi$ almost everywhere, so that $f$ is Lebesgue measurable with
$$
\int f \, d\lambda(x) = \int \psi \, d \lambda(x) = \lim_n \int \psi_n \, d\lambda = \lim \int \psi_n \, dx = \int f \, dx.
$$
This completes the proof.

For improper Riemann integrals, the claim is false however, as (cf. the answer by Peter) the example of $$\frac{\sin(x)}{x}$$ shows. 

The point here is that Lebesgue integrability of $f$ implies integrability of $|f|$, whereas for the (improper) Riemann integral it can happen that $f$ is integrable, although $|f|$ is not.

Finally, one can even show (using a variant of the proof above) that a bounded function $f : [a,b] \to \Bbb{R}$ is Riemann integrable if and only if the set of discontinuities of $f$ is a set of Lebesgue measure zero.
