This is impossible. Baire proved that if a function defined on $\mathbb R$ is of Baire class 1, then it is continuous everywhere except, possibly, for a meagre set. And by another Baire's theorem a complement of a meagre set in $\mathbb R$ is dense.
An elementary exposition of this and related results can be found in the nice little book by Oxtoby.
Edit. Another good reference which covers Baire's theorems and provides some historical background is "The calculus gallery" by W. Dunham.
Choose $\varepsilon > 0$ and consider sets defined by
$$\Sigma_k = \{E:\ \left|\int\limits_{E} (f_n-f_m) \right| \leqslant \varepsilon, \textrm{ if } n,m \geqslant k \}$$
Since for any measurable set a limit of integrals exists, we have
$\Sigma = \bigcup\limits_{k} \Sigma_k$. Note, that given an integrable function $f$, the functional $f(E):= \int\limits_{E}f$ is continuous respect to $E$ in metric $\delta$. Indeed, $f(E)-f(F) = \int f \cdot ( 1_E - 1_F )$, hence $|f(E)-f(F)| \leqslant \int |f| \cdot | 1_E - 1_F | = \int |f| \cdot 1_{E \Delta F} = \int\limits_{E\Delta F} |f|$.
The last expression tends to $0$ if $|E\Delta F|$ tends to $0$ because of integrability of $f$, equivalently if $d(E,F)\to 0$. This remark shows that sets $\Sigma_k$ are closed as an intersection of closed subsets of the space $(\Sigma,d)$.
From Baire theorem we obtain that one of sets $\Sigma_k$ has an interior point. Therefore, there exists a measurable set $E_0$ and integer $k$ such that the inequality
$ |f_n(E)-f_m(E) | \leqslant \varepsilon$ holds, whenever $|E\Delta E_0| \leqslant \delta$ and $m,n\geqslant k$. We will show, that this inequality holds in fact for any set $E$, provided that its measure is sufficiency small.
By identities $\mathbf{1}_{E\cup E_0} - \mathbf{1}_{E_0} = \mathbf{1}_{E\cap E_0^{c}}$ and $\mathbf{1}_{E_0}-\mathbf{1}_{E_0\setminus E} = \mathbf{1}_{E\cap E_0}$ we obtain for an arbitrary integrable $f$
$$f(E) = f(E \cap E_0^{c}) + f(E\cap E_0) = f(E\cup E_0) - f(E_0) + f(E_0) - f(E_0\setminus E)$$
If $|E| < \delta$, then all of sets $E_0,E_0\cup E, E_0\setminus E$, belong to the ball $\{E:\ |E\Delta E_0| < \delta \}$. Applying the last inequality to $f_n-f_m$ and $|E| < \delta$ we get
$$ |f_n(E)-f_m(E)| \leqslant 2\varepsilon \quad \textrm { if } |E|<\delta, \ n,m\geqslant k$$
Finally, observe that the finite family of integrable functions $f_1,\ldots, f_k$ is obviously locally uniformly integrable, i.e. $\sup\limits_{i\leqslant k} |f_i(E)| \to 0$ if $|E|\to 0$. Therefore, for sufficiency small $\delta'$ we have
$$|f_i(E)|\leqslant \varepsilon \quad \textrm{ if } |E| < \delta', i\leqslant k$$
Gluing together two estimates that have been derived, we see that for some positive $\delta$
$$ \left|\int\limits_{E} f_n\right| \leqslant 3\varepsilon, \quad \textrm{ if } |E| \leqslant \delta$$
We have estimated integrals for functions $f_n$, instead their modulus. It doesn't matter, however. Namely, applying the last estimate to the set $E\cap \{f_n > 0\}$ and $E\cap \{f_n < 0\}$ respectively (both contained in $E$ hence with a smaller measure), gives finally
$$ \int\limits_{E} |f_n| \leqslant 6\varepsilon, \quad \textrm{ if } |E| \leqslant \delta$$
What finished the proof.
Best Answer
For a proof of the much stronger result indicated above, see Page 53 here; the theorem states the following:
Let $f(x)$ be $C^\infty$ on $(c,d)$ such that for every point $x$ in the interval there exists an integer $N_x$ for which $f^{(N_x)}(x)=0$; then $f(x)$ is a polynomial.