Uniformly bounded sequence of functions Let $f(x)$ a continuous function on $\mathbb{R}$ such that the improper Riemann integral of $f$ on $\mathbb{R}$ is $1$, $\int_{-\infty}^{+\infty}f(x)dx=1$. 
For every $k \in \mathbb{N}$ we define $$f_k(x)=kf(kx),\qquad x \in \mathbb{R}.$$ Also, for every $k \in \mathbb{N}$ we set $$r_k(x)=\int_{-\infty}^{x}f_k(y)dy,\qquad x \in \mathbb{R}.$$ Integration by substitution leds to $$r_k(x)=\int_{-\infty}^{kx}f(z)dz$$ for every $k$ and $x$, so that $\{r_k(x)\}$ converges pointwise to $$H(x)=\begin{cases}1\qquad x>0\\0\qquad x<0\end{cases}$$ on $\mathbb{R}$. I'm in trouble proving that

$\{r_k(x)\}$ is uniformly bounded on $\mathbb{R}$, ie exist $M>0$ such that $|r_k(x)|\leq M$ for every $k$ and $x$.

Since with $k$ fixed we have $\lim_{x\to -\infty}r_k(x)=0$ and  $\lim_{x\to +\infty}r_k(x)=1$ it is simple to show that $r_k$ is bounded, but I cannot find a constant that bounds $r_k$ for every $k$. 
Any help would be really appreciated.
 A: Since the improper integral is convergent with $\int_{-\infty}^\infty f(x) \, dx = 1$, there exists finite $A$ and $A'$ such that $\int_0^\infty f(x) \, dx = A$ and $\int_{-\infty}^0 f(x) \, dx = A'$. 
Since $\lim_{x \to \infty} \int_0^x f(z) \, dz = A$, there exists $C > 0$ such that for all $x \geqslant C$ we have 
$$\left|\int_{0}^x f(z) \, dz  \right| - |A| \leqslant \left|\int_{0}^x f(z) \, dz- A\right| \leqslant \frac{1}{2}\\ \implies \left|\int_{0}^x f(z) \, dz  \right| \leqslant |A| +\frac{1}{2}= M'$$
Since $x \geqslant C $ implies that $kx \geqslant C$ for all $k \in \mathbb{N}$, it follows that
$$\left|\int_{0}^{kx} f(z) \, dz  \right| \leqslant M', \quad (x \geqslant C, k \in \mathbb{N})$$
Since $F(x) = \int_{0}^x f(z) \, dz$ continuous on the compact interval $[0,C]$, it follows that $F$ is bounded and there exists $M'' > 0$ such that for all $x \in [0,C]$ we have $\left|\int_{0}^{x} f(z) \, dz  \right| \leqslant M''$. 
Hence, if $x \in [0,C]$, then for all $k \in \mathbb{N}$,
$$\left| \int_{0}^{kx} f(z) \, dz  \right| \begin{cases} \leqslant M', &kx \geqslant  C \\ \leqslant M'', &0\leqslant kx < C   \end{cases}$$
Thus, for all $x \geqslant 0$ and $k \in \mathbb{N}$ we have 
$$\left| \int_{0}^{kx} f(z) \, dz  \right| \leqslant \max(M',M''),$$
and
$$\left| \int_{-\infty}^{kx} f(z) \, dz  \right| \leqslant \left| \int_{-\infty}^0 f(z) \, dz  \right| + \left| \int_{0}^{kx} f(z) \, dz  \right|= |A'| + \max(M',M'') := M_+ $$
We can make a similar argument to show that $\int_{kx}^0 f(z) \, dz$ is uniformly bounded for all $x < 0$ and $k \in \mathbb{N}$.  Since $\int_{-\infty}^{kx}f(z) \, dz  = A' - \int_{kx}^0 f(z) \, dz$ it follows there exists a uniform bound $M_-$ where $\left| \int_{-\infty}^{kx} f(z) \, dz \right| \leqslant M_-$
Therefore, for all $x \in \mathbb{R}$ and $k \in \mathbb{N}$ we have
$$\left|\int_{-\infty}^{kx} f(z) \, dz\right| \leqslant \max(M_+, M_-) := M $$
