Bounded (from below) continuous local martingale is a supermartingale

Here's my solution, based on a hint by @saz. I am not 100% percent it is true.

Let $M'(t) = M(t) + c$. Since $M'(t)$ is a (continuous) local martingale, there exists a sequence $T_n$ of stopping times such that $M'_n(t) = M'(t\wedge T_n)$ is a martingale for each $n$. That is, with respect to a filteration $\mathcal{F}(t)$, $t\ge 0$, it holds that for all $0\le s\le t$, $$\mathbb{E}\left(M'_n(t)\mid\mathcal{F}(s)\right)=M'_n(s).$$

By Fatou's lemma, \begin{eqnarray*} \mathbb{E}\left(\left|M'(t)\right|\right) &=& \mathbb{E}\left(M'(t)\right)\\ &=& \mathbb{E}\left(\liminf M'_n(t)\right)\\ &\le& \liminf\mathbb{E}\left(M'_n(t)\right) = \mathbb{E}\left(M'(0)\right) < \infty \end{eqnarray*}

By Fatou's lemma again, $$\mathbb{E}\left(M'(t)\mid\mathcal{F}(s)\right) =\mathbb{E}\left(\underline{\lim} M'_n(t)\mid\mathcal{F}(s)\right) \le \underline{\lim}\mathbb{E}\left(M'_n(t)\mid\mathcal{F}(s)\right) =\underline{\lim} M'_n(s) = M'(s)$$ and it follows that $M'$ is a supermartingale, hence $M$ is a supermartingale.

Note: I have edited my answer to include proof of integrability. While doing that, I figured out that one should assume $\mathbb{E}(M'(0))<\infty$. I'll update the question.

Hint Apply Fatou's lemma in order to prove $(M_t)_{t \geq 0}$ a supermartingale.