Let there be a n-sized sample (iid r.v.'s), $\{X_{n}\}_{n \geq 1}$, from a r.v. $X \sim \operatorname{Poisson}(\lambda)$. Knowing that $\bar{X} \xrightarrow[]{P} \lambda$, find a function $T(.)$ for which $T(X_{1},...,X_{n}) \xrightarrow[]{P} e^{\lambda}$ holds. Also find a $G(.)$ for which $G(X_{1},...,X_{n}) \xrightarrow[]{P} \mathbb{P(X=0)}+\mathbb{P(X=1)}$ holds.
My thoughts are:
We know that \begin{align} \lim_{n \xrightarrow[]{} \infty} P(\lvert \bar{X}-\lambda\rvert > \epsilon) &=0 \\ \mathbb{P}(X=0)+\mathbb{P}(X=1) &= e^{-\lambda} + \lambda e^{-\lambda} \end{align} And the conditions for the desired convergences are: \begin{align} \lim_{n \xrightarrow[]{} \infty} P(\lvert T(X_{1},...,X_{n})-e^{\lambda}\rvert > \epsilon) &=0\\ \lim_{n \xrightarrow[]{} \infty} P(\lvert G(X_{1},...,X_{n})-(e^{-\lambda} + \lambda e^{-\lambda})\rvert > \epsilon) &=0 \end{align}
I'd like to know if we can choose: \begin{align} T(X_{1},...,X_{n}) = \bar{X}-\lambda +e^{\lambda} \\ G(X_{1},...,X_{n}) = \bar{X}-\lambda +e^{-\lambda} + \lambda e^{-\lambda} \end{align}
Because then the limits would turn into the first one that we know it's equal $0$, fulfilling the conditions.
|EDIT|
I found the following theorem:
The Continuous Mapping Theorem:
Let $Xn$ and $X$ be random variables and $g$, a continuous function.
If $Xn \xrightarrow[]{P} X$, then $g(Xn) \xrightarrow[]{P} g(X)$.
Therefore it would be simply:
\begin{align} T(X_{1},...,X_{n}) = e^{\bar{X}} \\ G(X_{1},...,X_{n}) = \lambda e^{\bar{X}} + e^{\bar{X}} \end{align}
Is this a fine solution? Thanks in advance.