Is the function $\displaystyle\liminf_{t\to t_0}f_t(x)$ Lebesgue measurable? 
Let $\{f_t:[0,1]\to \mathbb R \  |t\in I\}$, where $I\subset \mathbb R$ is an interval, be a class of Lebesgue measurable functions. Then is it true that the function $\displaystyle\liminf_{t\to t_0}f_t(x):[0,1]\to \mathbb R$ also Lebesgue measurable, where $t_0\in I$?

What I know is that if $f_1,f_2,...,f_n,...$ are measurable functions then $\displaystyle \liminf_{n\to \infty}f_n$ is also measurable.
For the above question, I was thinking that 
Let $\{x_n:n\in \mathbb N\}\subset I$ be a sequence from $I$ such that $\displaystyle \lim_{n\to \infty}x_n=t_0$. Then $\displaystyle\liminf_{t\to t_0}f_t(x)=\liminf_{n\to \infty}f_{x_n}(x)$ from the definition. And the function $\displaystyle\liminf_{n\to \infty}f_{x_n}(x)$ is measurable. So the function $\displaystyle\liminf_{t\to t_0}f_t(x)$ is measurable.
I don't know it is true or not.
It would be very helpful if anyone checks the above. 
Thank you.
 A: No, this is not measurable.
For a counterexample, we choose $I=[0,1]$ and $t_0=0$.
Let $A\subset[0,1]$ be a non-measurable set.
We define $f_0=0$ and
$$
f_t = -\chi_{\{\sin^2(1/t)\}\cap A}
\qquad t>0.
$$
Then $f_t$ is measurable.
However, it can be shown that $\liminf_{t\to t_0} f_t=-\chi_A$,
which is non-measurable.
A: An example.  
Let's use $t_0 = 0$ so we are taking liminf as $t \to 0$.
Let $E \subset \mathbb R$ be a non-measurable set of cardinal $2^{\aleph_0}$.
The liminf funcion $f$ will the the characteristic function of the complement of $E$.
$$
f(x) = \begin{cases}
0,& x \in E
\\
1,& x \not\in E
\end{cases}
$$
So $f$ will be non-measurable.
Each function $f_t$ will be $1$ except for one point, where it is $0$.
$$
f_t(x) = \begin{cases}
0, &x = u_t\\1, &\text{otherwise}
\end{cases}
$$
so $f_t$ is measurable.
The points $u_t$ remain to be chosen.
Divide $(0,1]$ into invervals
$$
I_n = \left(\frac{1}{n+1},\frac{1}{n}\right]
$$
For each $n$, let the points $u_t$ be chosen so that
$$
\{u_t: t \in I_n\} = E
$$
Both $I_n$ and $E$ have cardinal $2^{\aleph_0}$.
Finally, we have $\liminf_{t \to 0}f_t(x) = f(x)$.  Indeed, if $x \not\in E$, then $f_t(x) = 1$ for all $t$.  And if $x \in E$, then $f_t(x) = 0$ for infinitely many $t$ approaching $0$ and $1$ for all other $t$.
