Let $(X,\mathcal{A},\mu)$ be a measureable space, and assume that $\mu(X)<\infty$. Let $\left \{ u_{n} \right \}_{n\geq 1}$ be a sequence of functions in $\mathcal{L}^{1}(\mu)$ that converges uniformly to a function $u:X\to\mathbb{R}$.
a) Show that $u\in \mathcal{L}^{1}(\mu)$.
b) Show that $$\lim_{n\to\infty}\int_{X}u_{n}\, \mathrm{d}\mu=\int_{X}u\, \mathrm{d}\mu.$$
Could you please correct my answers and help me to answer if I am wrong.
My answers:
a) Because of the convergence we have for all $\epsilon>0$ there exists $N\in \mathbb{N}$ such that $|u(x)-u_{n}(x)<\epsilon$ whenever $n\geq N$ for all $x\in X$. We see that \begin{align*} \left | \int_{X} u\, \mathrm{d}\mu \right |&=\left | \int_{X} u-u_{n}\, \mathrm{d}\mu+\int_{X} u_{n}\, \mathrm{d}\mu \right | \\ &\leq\int_{X}\left | u-u_{n} \right |\, \mathrm{d}\mu+\int_{X} \left |u_{n} \right |\, \mathrm{d}\mu \\ &<\epsilon\int_{X}\,\mathrm{d}\mu+ \int_{X} \left |u_{n} \right |\, \mathrm{d}\mu \end{align*} which clearly shows that $u\in \mathcal{L}^{1}(\mu)$ because $\int_{X}\,\mathrm{d}\mu=\mu(X)<\infty$ and $\left \{ u_{n} \right \}_{n\geq 1}\subset \mathcal{L}^{1}(\mu)$.
b) We have for all $\epsilon >0$ $$ \left | \int_{X}u_{n}\, \mathrm{d}\mu-\int_{X}u\, \mathrm{d}\mu \right |\leq \int_{X}\left | u_{n}-u \right |\, \mathrm{d}\mu<\epsilon\mu(X)<\infty $$ which shows that the limit exists.
With (a) you don't know ahead of time that $u$ is integrable so it is a good idea to consider only nonnegative functions. Since $|u| \le |u - u_n| + |u_n|$ you have $$ \int_X |u| \, d\mu \le \int_X |u - u_n| \, d\mu + \int_X |u_n| \, d\mu < \epsilon \mu(X) + \int_X |u_n| \, d\mu < \infty$$ provided that $|u - u_n| < \epsilon$. Thus $u \in L^1(\mu)$.
Your solution for (b) is off a bit. Choose $N$ so that $n \ge N$ implies $|u_n(x) - u(x)| < \dfrac{\epsilon}{\mu(X)}$ for all $x \in X$ (we ignore the trivial possibility that $\mu(X) = 0$). Then according to your calculation, $$ n \ge N \implies \left| \int_X u_n \, d\mu - \int_X u \, d\mu \right| < \epsilon. $$