Let $f: \mathbb{R}^n \to \mathbb{R}$ be a bounded below function and let $X \in \mathbb{R}^n$ be a random vector. Is $\mathbb{E}_{X}[f(X)]$ bounded below?
My try
$X$ is a random vector that is distributed by some unknown distribution $\mathcal{D}$. Thus, one can write the following:
$$ \mathbb{E}_{X}[f(X)] = \int_{x_1,\dots,x_n}f(x)dx_1\dots dx_n $$ how can we show that the above integral is bounded below?
For clarity, let's relabel your function. Let $h:\mathbb{R}^n \to \mathbb{R}$ be bounded below, and let $X \in D$ be distributed according to some pdf $f$. In particular, $f \ge 0$ and $\int_D f(x)\ dx = 1$.
Since $h$ is bounded below, then $h(x) \ge M$ for all $x$, thus, $$ \mathbb{E}[h(X)] = \int_D h(x) f(x)\ dx \ge \int_D M f(x)\ dx = M \int_D f(x)\ dx = M $$ where the integral inequality is valid since $f(x) \ge 0$.
UPDATE
Following the suggestion of @Adayah below, note the above proof only works for continuously distributed random vectors, but if $\mu$ is a general distribution, we can generalize the argument by $$ \mathbb{E}[h(X)] = \int_D h(x) \ d\mu(x) \ge \int_D M \ d\mu(x) = M \int_D \ d\mu(x) = M $$