We need to prove the following lemma:
Lemma: If $f$ is non-negative and Riemann integrable on $[a, b]$ with $$\int_{a}^{b}f(x)\,dx > 0$$ then there is a sub-interval of $[a, b]$ on which $f$ is positive.
The lemma is proved easily by using the definition of Riemann integral. Let $$I = \int_{a}^{b}f(x)\,dx > 0$$ and let $0 \leq f(x) < M$ for all $x \in [a, b]$. By definition of Riemann integral there is a partition $$P = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\}$$ of $[a, b]$ such that the Riemann sum $$S(f, P) = \sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1}) > \frac{I}{2}$$ Let $$\epsilon = \frac{I}{2(M + b - a)}$$ and consider the set $J = \{x: x \in [a,b], f(x) \geq \epsilon\}$. Now we split the Riemann sum $S(f, P)$ as $$S(f, P) = \sum_{i \in A}f(t_{i})(x_{i}- x_{i - 1}) + \sum_{i \in B}f(t_{i})(x_{i}- x_{i - 1})$$ where $$A = \{i: [x_{i - 1}, x_{i}] \subseteq J\}, B = \{i: i \notin A\}$$ Now for $i \in A$ we can see that $f(t_{i}) \leq M$ and for $i \in B$ we can choose $t_{i}$ such that $f(t_{i}) < \epsilon$. Then we can see that $$\epsilon(M + b - a) = \frac{I}{2} < S(f, P) < M\sum_{i \in A}(x_{i} - x_{i - 1}) + \epsilon(b - a)$$ It now follows by simple algebra that $$\sum_{i\in A}(x_{i} - x_{i - 1}) > \epsilon$$ and therefore there exist intervals of type $[x_{i - 1}, x_{i}]$ where $i \in A$ where $f(x) \geq \epsilon$. The above proof is taken from my favorite book Mathematical Analysis by Tom M. Apostol.
Now we need to make use of this lemma for solving the current problem. Let's assume that $$\lim_{n \to \infty}\int_{0}^{1}f_{n}(x)\,dx = \lim_{n \to \infty}I_{n} = c > 0$$ Then it follows that after a certain value of $n$ we have $I_{n} > c/2$. And by our lemma it follows that $f_{n}(x) \geq d > 0$ where $d$ is some constant dependent only on $c$ and not on $n$. Moreover we see that there is a partition $P_{n}$ of $[0, 1]$ such that $f_{n}(x)\geq d$ in some of the sub-intervals made by $P_{n}$ (whose total length also exceeds $d$) and $f_{n}(x) < d$ for some point in remaining sub-intervals. If necessary we can replace $P_{n + 1}$ by $P_{n} \cup P_{n + 1}$ and hence it is OK to assume that $P_{n + 1}$ is a refinement of $P_{n}$.
Since $f_{n}$ is monotonically decreasing it follows that the sub-intervals of partition $P_{n + 1}$ where $f_{n + 1}(x) \geq d$ are contained in the corresponding sub-intervals of $P_{n}$ and again these sub-intervals of $P_{n + 1}$ have a total length greater than $d$. Thus it follows that there is a sub-interval of $[0, 1]$ where $f_{n}(x) \geq d > 0$ for all $n$ after a certain value. This contradicts the fact that $f_{n}$ converges pointwise to $0$.
The above proof is wrong and the wrong inferences are striked out. The approach however can be salvaged with some more effort. An alternative approach is given in this answer.
The integral and the limit can indeed be interchanged.
To see this, first note that if $g \in L^2_{\mathrm{loc}}(\Bbb{R})$ is $2\pi$-periodic, then
\begin{align*}
\int_{\Bbb{R}}
g(x) \, \varphi(x)
\, d x
& = \sum_{k \in \Bbb{Z}}
\int_{2\pi k}^{2\pi k+2\pi}
g(x) \, \varphi(x)
\, d x \\
& = \sum_{k \in \Bbb{Z}}
\int_{0}^{2\pi}
g(x) \, \varphi(x + 2\pi k)
\, d x \\
& = \int_0^{2\pi}
g(x) \, P \varphi (x)
\, d x,
\end{align*}
where $P \varphi(x) = \sum_{k \in \Bbb{Z}} \varphi(x + 2\pi k)$ is bounded
(since $\varphi$ is a Schwartz function) and $2\pi$-periodic.
The interchange of the series and the integral above can be easily justified using the dominated
convergence theorem.
Now, define $f_N := \frac{1}{2\pi} \sum_{k=-N}^N b_k e^{i k x}$, and note that
$f_N \to f$ in $L^2([0,2\pi])$.
Since $P \varphi \in L^\infty([0,2\pi]) \subset L^2 ([0,2\pi])$, we see
\begin{align*}
\int_{\Bbb{R}}
f_N (x) \, \varphi(x)
\, d x
& = \int_0^{2\pi}
f_N (x) \, P \varphi (x)
\, d x \\
& \xrightarrow[N\to\infty]{} \int_0^{2\pi} f(x) \, P \varphi (x) \, d x \\
& = \int_{\Bbb{R}}
f(x) \, \varphi(x)
\, d x ,
\end{align*}
which is what you wanted to show.
Best Answer
The sequence $(f(nx))$ converges to the function $h$ defined by $$h(x)=\left\{\begin{array}{cl}\\L&\quad\text{if}\ 0<x\leq1\\ f(0)&\quad\text{if}\ x=0 \end{array}\right.$$ and since the continuous function has a finite limite at $+\infty$ then it's bounded on the interval $[0,+\infty)$ so there's $M>0$ s.t. $$|f(nx)|\leq M \quad \forall\ 0\leq x\leq 1$$ hence we apply the dominated convergence theorem and we have $$\lim_{n\to\infty}\int_0^1f(nx)dx=\int_0^1h(x)dx=L.$$