Showing $\frac{\sin x}{x}$ is NOT Lebesgue Integrable on $\mathbb{R}_{\ge 0}$

You might want to start by showing that $f$ is conditionally convergent. This would show that $f$ is not Lebesgue integrable. Use the fact that $$\int_{(n - 1)\pi}^{n\pi} \frac{\left| \sin x \right|}{x} dx \geq \frac{1}{n\pi}\int_{(n - 1)\pi}^{n\pi} \left| \sin x \right| dx = \frac{2}{n\pi}$$ to get $$\int_0^{N\pi} \frac{\left| \sin x \right|}{x} dx \geq \frac{2}{\pi}\sum_{n = 1}^N \frac{1}{n}$$ which shows that $$\int_0^\infty \frac{\left| \sin x \right|}{x} dx$$ is divergent.

By definition, a measurable function $f=f(x)$ is Lebesgue integrable over a measurable set $E$ if and only if the Lebesgue integrals$\int_E f^+(x)\,dx$ and $\int_Ef^-(x)\,dx$ are both finite. Let us look at $f^+(x)$, which in this case is $\sin^+(x)/x$. Since $f^+(x)\geqslant 0$, $f^+(x)$ is integrable if and only if $\int f^+(x)\,dx<\infty$. Again, since $f^+(x)\geqslant 0$, we can compute its Lebesgue integral with the Monotone Convergence Theorem, and then estimate it from below: \begin{align*} \int_0^\infty f^+(x)\,dx &= \int_0^\infty \frac{\sin^+(x)}{x}\,dx \\ &\color{red}{=}\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}{\color{green}{x}}\,dx & \color{red}{\text{Monotone Convergence Theorem}} \\ &\color{green}{\geqslant}\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}{\color{green}{(2k+1)\pi}}\,dx \\ &= \sum_{k=0}^\infty\frac{1}{(2k+1)\pi}\color{blue}{\int_{2k\pi}^{(2k+1)\pi}\sin(x)\,dx} \\ &\color{blue}{=} \sum_{k=0}^\infty\frac{\color{blue}{2}}{(2k+1)\pi} \\ &= \sum_{k=0}^\infty\frac{1}{(k+1/2)\pi} \end{align*} By the Limit Comparison Test applied to the series obtained above and the harmonic series, we conclude that the series diverges, hence $\int_0^\infty f^+(x)\,dx = \infty$, so $f(x)$ is not integrable.

Let $f(x)=\frac{\sin x}{x}$. Surely the function is measurable, so we need to prove that $$\int|f(x)|dx=+\infty$$ In order to do that we need to look closer to the function. To this end, note that it is almost like $x\mapsto\frac{1}{x}$, however our function has zeros so we cannot compare them directly. A better approach is to use that the local maximums are attained at $x=\frac{\pi}{2}+\pi \cdot n$, in fact we have $$|f(x)| \geq \frac{1}{\sqrt{2}x}, \quad\text{for $ \left|\frac{(2n+1)\pi}{2}-x\right|\leq \frac{\pi}{4}$}$$ Now, do you see how to estimate the function from below?