Weak Law of Large Numbers -Special Case For independent and identically distributed random variables $X_1, X_2, \cdots$ with finite mean and variance, by the weak Law of Large Numbers, $\frac{1}{N}\sum_{i=1}^{N} X_i$ converges in probability to $\mathbb{E}[X_i]$:
$$\frac{1}{N}\sum_{i=1}^{N} X_i \xrightarrow{p}\mathbb{E}[X_i]$$

If $f(N)$ is a continuous function such that $\lim_{N \to \infty} f(N)=c \in \mathbb{R}^{+}$, then is the following true?
  $$\frac{f(N)}{N}\sum_{i=1}^{N} X_i \xrightarrow{p}c \mathbb{E}[X_i]$$

We know that $\lim\limits_{N \to \infty} \mathbb{P}\left(|\frac{1}{N}\sum_{i=1}^{N} X_i - \mathbb{E}[X_i]|>\epsilon\right)=0$ and thus
$$\lim\limits_{N \to \infty} \mathbb{P}\left(|\frac{f(N)}{N}\sum_{i=1}^{N} X_i - f(N)\mathbb{E}[X_i]|>\epsilon f(N)\right)=0$$
How can we continue form here?
 A: We can assume that $|f(N)|\le M$, for all $N$ and certain $M>0$. Moreover, there exist $N_0$ such that, for $N>N_0$, $|f(N)-c|< \varepsilon/(2 E(X_1))$. Since
\begin{align*}
\Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big| \le \Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - f(N)E(X_1) \Big| + \big|f(N)-c\big|\, E(X_1),
\end{align*}
we can have that
\begin{align*}
\Big|\frac{1}{N}\sum_{i=1}^N X_i - E(X_1) \Big| &\ge \frac{\Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big| - \big|f(N)-c\big|\,E(X_1)}{f(N)}\\
&\ge \frac{\Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big| - \frac{\varepsilon}{2}}{M}.
\end{align*}
Then, for $N> N_0$,
\begin{align*}
\left\{\omega: \Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big|\ge \varepsilon\right\} \subset \left\{\omega: \Big|\frac{1}{N}\sum_{i=1}^N X_i - E(X_1) \Big|\ge \frac{\varepsilon}{2M}\right\}.
\end{align*}
That is,
\begin{align*}
\lim_{N\rightarrow\infty}\mathbb{P}\left(\Big|\frac{f(N)}{N}\sum_{i=1}^N X_i - cE(X_1) \Big|\ge \varepsilon\right) = 0.
\end{align*}
