Finding a consistent Estimator for $\mathbb{E}(X^2)$ Given $X_1,\dots,X_n\stackrel{iid}{f}(x;\mu,\sigma^2)$ with population mean $\mu$ and population standard deviation $\sigma$. I want to find a consistent estimator for $\mathbb{E}(X^2)$.
By the variance formula we know $\mathbb{E}(X^2)=\mathrm{Var}(X_i)+[\mathbb{E}(X_i)]^2=\sigma^2+\mu^2$
So if I am not mistaken $\sigma^2+\mu^2$ is our estimand and we want to find an estimator that will converge to this estimand as $n\rightarrow\infty$. $\color{red}{\text{correct me if im wrong!}}$
We know $\bar{X}\stackrel{p}{\longrightarrow}\mu$ , I am assuming squaring will preserve continuity so $\bar{X}^2\stackrel{p}{\longrightarrow}\mu^2$.  $\color{red}{\text{correct me if im wrong!}}$
But I do not know how to proceed from here. It does not feel right to say $\mathbb{E}(\bar{X}^2)\stackrel{p}{\longrightarrow}\mu^2+\sigma^2$. I am assuming I will have to show it converges using chebyshev's.
I know in general $\mathbb{E}(\bar{X}^2)=\frac{\sigma^2}{n}+\mu^2$, so I was considering $n\bar{X}$ $\color{red}{\text{ (is this a good idea?)}}$
Any help or guidance would be appreciated!

If you want to estimate $\mathbb{E}[X^2]$, a natural estimator would be to simply take the sample mean of $X_i^2$. Then by the weak law of large numbers, $\frac{1}{n} \sum^n_{i=1} X_i^2 \stackrel{p}{\rightarrow} \mathbb{E}[X^2]$.

If you want to use your approach (which seems to also work when pushed a bit), a useful tool would be the continuous mapping theorem. Thus we know $\overline{X} \stackrel{p}{\rightarrow} 
\mu$, and so since squaring is continuous, indeed $\overline{X}^2 \stackrel{p}{\rightarrow} \mu^2$. For the variance, we can use the usual sample variance estimator, i.e. $\frac{1}{n} \sum^n_{i=1} (X_i - \overline{X})^2$, which converges to $\sigma^2$ in probability. Then the sum of these two estimators will also converge to $\mu^2 + \sigma^2$ in probability (again, by the continuous mapping theorem).
