$\int_{0}^{\infty}\frac{\sin(x)}{x}dx$ is riemann integrable.
But how to prove that it's not Lebesgue integrable ?
(I tried contradiction).
If $f(x)=\frac{\sin(x)}{x}$, I supppose that $f\in L^1(\mathbb{R^+})$.
then, I' ll define the function sequence : $f_n(x)=\frac{\sin( x)}{x}\mathbb{1}_{[0,n]}(x).$
We have $f_n$ converges $\lambda$-a.e to $f$.
and $|f_n|\leq|f|$, for all $1\leq n$.
and $f$ integrable (hypothesis).
then with the dominated convergence theorem
$$\lim_n\int_{0}^{n}f_nd\lambda=\int_{\mathbb{R^+}}fd\lambda$$
which is equivalent to $$\int_{0}^{\infty}\frac{\sin(x)}{x}dx=\int_{\mathbb{R^+}}f d\lambda$$
Thats basically what I did.
Hint. If such function is Lebesgue integrable then also its absolute value is Lebesgue integrable. Let $n$ a positive integer, then $$ \int_0^{2\pi n} \frac{|\sin x|}{x} \,dx = \sum_{k=0}^{n-1} \int_{2\pi k}^{2\pi(k+1)} \frac{|\sin x|}{x}\,dx \ge \sum_{k=0}^{n-1} \frac{1}{2\pi (k+1)} \int_{0}^{2\pi} |\sin x|\,dx $$ Can you take it from here?