To respond to your comment on Byron's answer:
The functional monotone class theorem is a very useful result and well worth knowing. However, you can also get this result with arguments that may be more familiar. To recap, we want to show:
Suppose $\mu', \mu''$ are two probability measures on $\mathbb{R}$, and we have $\int f\,d\mu' = \int f\,d\mu''$ for all bounded continuous $f$. Then $\mu' = \mu''$.
One could proceed as follows:
Exercise. For any open interval $(a,b)$, there is a sequence of nonnegative bounded continuous functions $f_n$ such that $f_n \uparrow 1_{(a,b)}$ pointwise.
(For example, some trapezoidal-shaped functions would work.)
If $f_n$ is such a sequence, we have $\int f_n \,d\mu' = \int f_n \,d\mu''$ for each $n$. By monotone convergence, the left side converges to $\int 1_{(a,b)}\,d\mu' = \mu'((a,b))$ and the right side converges to $\mu''((a,b))$. So $\mu'((a,b)) = \mu''((a,b))$, and this holds for any interval $(a,b)$.
Now you can use Dynkin's $\pi$-$\lambda$ lemma, once you show:
Exercise. The collection
$$\mathcal{L} := \{B \in \mathcal{B}_\mathbb{R} : \mu'(B) = \mu''(B)\}$$
is a $\lambda$-system. (Here $\mathcal{B}_{\mathbb{R}}$ is the Borel $\sigma$-algebra on $\mathbb{R}$.)
We just showed that the open intervals are contained in $\mathcal{L}$. But the open intervals are a $\pi$-system which generates $\mathcal{B}_{\mathbb{R}}$. So by Dynkin's lemma, $\mathcal{B}_\mathbb{R} \subset \mathcal{L}$, which is to say $\mu' = \mu''$.
Here $g_{n}\left(\omega\right)=\sup\left\{ f_{m}\left(\omega\right)\mid m\geq n\right\} $
for each $\omega\in\Omega$ right?
Then $g_{n}\left(\omega\right)\leq a$
if and only if $f_{m}\left(\omega\right)\leq a$ for each $m\geq n$.
You could also say that $\omega\in\left\{ g_{n}\leq a\right\} $ if
and only $\omega\in\cap_{m\geq n}\left\{ f_{m}\leq a\right\} $.
This is true for every $\omega\in \Omega$ so that
means exactly that the sets $\left\{ g_{n}\leq a\right\} $ and $\cap_{m\geq n}\left\{ f_{m}\leq a\right\} $
are the same.
It shows that set $\left\{ g_{n}\leq a\right\} $ is a countable intersection
of measurable sets, and this implies that the set is measurable.
addendum:
Characteristic for a $\sigma$-algebra is that it is closed under complements and countable unions. Then it must also be closed under countable intersections. This because: $$\cap_{n=1}^{\infty}A_{n}=\left(\cup_{n=1}^{\infty}A_{n}^{c}\right)^{c}$$
Best Answer
Spelling out geetha290krm's comment; we have (evaluating pointwise) $$ (\inf_{r\in\Bbb R}f_r)(\omega) = \inf_{r\in\Bbb R}(f_r(\omega)) = \inf_{r\in\Bbb R}(\varphi_\omega(r)) = \inf_{q\in\Bbb Q}(\varphi_\omega(q)) = \inf_{q\in\Bbb Q}(f_q(\omega)) = (\inf_{q\in\Bbb Q}f_q)(\omega), $$ so $$\inf_{r\in\Bbb R}f_r = \inf_{q\in\Bbb Q}f_q.$$
For passing from $\Bbb R$ to $\Bbb Q$, see Supremum over real interval equals supremum over rational 'interval' for continuos functions?. To finish from the above, see The Supremum and Infimum of a sequence of measurable functions is measurable.