$f\in L^1(\mathbb{R}^N)$ and Lipschitz continuous, then $f\in L^\infty(\mathbb{R}^N)$.

Question

$f\in L^1(\mathbb{R}^N)$ and Lipschitz continuous, then $f\in L^\infty(\mathbb{R}^N)$.

Denote the Lipschitz constant of $f$ as $C$, suppose $f$ is not bounded a.e., then $\mu(\{f> k+C)\}) > 0 $ for all $k\in \mathbb{N}$. Take $x_k\in \{f> k+C)\}$, we have $$\int_{\mathbb{R}^N} |f| dx \geq \int_{B(x_k,1)} |f| dx \geq \int_{B(x_k,1)} k\;dx = k\alpha(N)$$ where $\alpha(N)$ is the volume of the unit ball.

The last inequality is from the following, since $f(x_k) \geq k+C$, then for each $y\in B(x_k,1)$, we have $$|f(x_k) - f(y)|\leq C |x_k - y| \leq C, $$ therefore $|f| \geq k$ on $B(x_k, 1)$.

This gives us a contradiction since $k\alpha(N)$ is not bounded.

Is this proof correct?

Thank you very much!