$Proposition$: Suppose that $\left\{ f_k \right\}^{\infty}_{k=1}$ is a sequence of Riemann integrable functions on the interval$[0,1]$ such that:
$$\int_{0}^{1}|f_k(x)-f(x)|dx \rightarrow 0 \, \, \, as \, k \rightarrow \infty$$
In summary we need to show that $f_k(n)-f(n)$ uniformly in $n$ as $k$ $\rightarrow$ $\infty$
$Proof$:
$Theorem \, 7.15.$: Suppose $f_n \in R[a,b]$ for all $n \in N$. Furthermore, suppose ${f_n}_n \in N$ converges uniformly to $f$ on $[a,b]$. Then $f \in R[a,b]$ therefore in (7.3)
$$(7.3)$$ $$\lim_{x \to +\infty}\int_{a}^{b}f_n(x)dx=\int_{a}^{b} \lim_{x \to +\infty}f_n(x)dx$$
$Lemma \, \, \, 1.1$: Suppose $f_k \in [0,1] for \, all \, n \in \mathbb{N}$ suppose $f_k$ converges to $f$ on $[0,1]$. Then $f$ $\in$ $R[0,1]$ from futher application of $Theorem \, \,7.15$ yields the following conclusions in $(7.4)$: $$(7.4)$$$$\lim_{k\to +\infty}\int_{0}^{1}|f_k(n)-f(n)|dx=\int_{0}^{1} \lim_{k \to +\infty}|f_n(n)-f(n)|dx$$
$Remark$: Considering our original proposition, the one can observe particularly within $Lemma \,\, 1.1$ that the initial functions taken with respect to our limit have been "swapped", this is initially done due to the fact we need to show that:$f_k(n)-f(n)$ uniformly in $n$ as $k \rightarrow \infty$.
$Lemma \, 1.2)$: From $(7.4)$ one can assume $n \subset \mathbb{R}$ and conduct the following manipulations in $(7.5)$ $$(7.5)$$ $$\int_{0}^{1}\lim_{k\to +\infty}|f_k(n)-f(n)|dx=\int_{0}^{1}\lim_{k\to +\infty}|f_k(n)| \, - \lim_{k\to +\infty}|f(n)|dx$$ $$\int_{0}^{1}(0-0)$$ $$\int_{0}^{1}(0)=0$$
This concludes our proof since, we've shown that:$\,$$f_k(n)-f(n)$ uniformly in $n$ as $k$ $\rightarrow$ $\infty$
What you seem to be trying to prove is false. Convergence in $L^1([0,1])$ does not imply uniform convergence, nor even pointwise convergence.
As a counterexample consider the sequence of Riemann integrable functions
$$f_1 = 1_{[0,1]}, f_2 = 1_{[0,1/2]}, f_3 = 1_{[1/2,1]}, f_4 = 1_{[0,1/4]}, f_5 = 1_{[1/4,1/2]}, \ldots$$
where
$$1_{[a,b]}(x) = \begin{cases}1, \,\, x \in [a,b]\\ 0, \,\, \text{otherwise} \end{cases}$$
Then
$$\int_0^1 |f_k(x) - 0| \, dx \to 0,$$
but $f_k(x)$ fails to converge to $0$ for any $x \in [0,1].$ Given $x$, for any $k$ there exists $k_1 > k$ such that $f_{k_1}(x) = 1.$