An strange multiplication of measurable functions Let $I$ be an uncountable (directed) set. Let $\{f_i\}_{i\in I}$ and $\{g_i\}_{i\in I}$ be two set of complex valued measurable functions on a measurable space $\Omega$. 
For every $t\in \Omega$, we let  $I_t$ given by 
$$I_t:=\{i\in I: both~~ f_i(t)~~and ~g_i(t)~~ are~~ non~~ zero\} $$
1-Assume for every $t\in \Omega$, the set $I_t$ is countable.
2-Assume that the map $\phi$ given by $\phi(t)=\sum_{I_t} f_i(t)g_i(t)$ is well defined. 
The question is this: Is $\phi$ measurable as well? 

In general, no, because of the following construction/special case:
Let $I\subset \Omega $ be arbitrary. Define $$f_i  (t)=\delta_i  (t)=\begin {cases}1&\text { if } i=t \\0,&\text  { else }.\end {cases} $$
and $g_i (t)=1$ for all $t $.
Then $I_t =\{t\} $ if $t \in I $ and $I_t =\emptyset  $ else is countable (even finite) for all $t $ and it is easy to verify
$$
\phi (t)=1_I  (t)=\begin {cases}1&\text { if } t\in I\\0&\text { else}\end {cases}
$$
Thus, if $I $ is not measurable, then $\phi $ is neither.
