Consider $E=(-3,3)$, and for $n=1,2, \ldots$ define the functions
\begin{align}h_n(x) :=
\begin{cases}
-2n - n^2x, \; \text{ if } x \in \left[-\frac{2}{n}, \,-\frac{1}{n} \right] , \\
n^2x , \; \text{ if } x \in \left[-\frac{1}{n} , \, \,\frac{1}{n}\right] , \\
2n - n^2x , \; \text{ if } x \in \left[\frac{1}{n} , \,\frac{2}{n}\right], \\
0, \; \text{ if } x \in \left(-3, -\frac{2}{n}\right) \cup \left(\frac{2}{n}, 3 \right). \end{cases}
\end{align}
Notice $\lim\limits_{n\to \infty} h_n(x)=0$ and $\lim\limits_{n\to \infty}\int\limits_{E} h_n dm=0$. Let $\delta>0$ be given. By the Archimedean Property, we may find a positive integer $N$ such that $N \delta > 4$. So we have that $m\left( \left[-\frac{2}{N}, \frac{2}{N} \right]\right)< \delta$ but notice $\int_{-\frac{2}{N}}^{\frac{2}{N}} \left| h_n \right| dm=2$ whenever $n \geq N$*. Hence the sequence of functions $\{h_n\}_{n=1}^\infty$ is not uniformly integrable.
*Note: Technically it suffices to observe that $\int_{-\frac{2}{N}}^{\frac{2}{N}} \left| h_N \right| dm=2$ since
$\neg \left(\forall \varepsilon\left(\varepsilon>0 \implies \exists \delta
\left( \delta>0 \land \forall n \forall A \left(m(A)<\delta \implies \int_A |\,h_n| dm < \varepsilon \right)\right) \right) \right)$
$\iff \exists \varepsilon \left( \varepsilon>0 \land \forall \delta \left( \delta>0 \implies \exists N \exists A \left(m(A)<\delta \land \int_A |\,h_N| dm \geq \varepsilon\right)\right) \right)$.
If $U$ is open in $\mathbb R$ then $f^{-1}(U)$ is open $M$ which means there is an open set $V$ in $\mathbb R^{n}$ such that $f^{-1}(U)=M\cap V$. This makes $f^{-1}(U)$ a measurable set. Hence $f$ is measurable. Integrability is not true in general. If $M=\mathbb R$ and $f(x)=\frac 1 {1+|x|}$ then $f$ is not integrable.
If $M$ has finite Lebesgue measure and $f$ is bounded by $C$ then $\int|f| \leq C\lambda (M) <\infty$ where $\lambda$ is the Lebesgue measure, so $f$ is Lebesgue integrable.
Best Answer
First this theorem is wrong
Consider $E = \mathbb R$, $f=0$ and $f_n = -\chi_{[n,n+1]}$. Where $\chi_{[n,n+1]}$ is the indicator function of the interval $[n,n+1]$.
They satisfy the hypothesis, but the conclusion doesn't hold as
$$ -1 = \int_{\mathbb R} f_n \neq \int_{\mathbb R} f = 0$$ while $(f_n)$ converges pointwise to the always vanishing function $f$.
Second this would more related to dominated convergence theorem
See Dominated convergence theorem if the hypothesis would be correctly set.