What can we say if we have a filter instead of an ultrafilter in Los's theorem? (Łoś's Theorem). Let $\mathcal{L}$ be a  language, $\{\mathcal{M}_i\}_{i\in I}$ a collection of  $\mathcal{L}$-structures  and $\mathscr{U}$ an ultrafilter over $I$. Let $\phi(x_1\dots,x_n)$ be an $\mathcal{L}$-formula and $g_1,\dots, g_n \in \prod_{i\in I} M_i$. Then, $\mathcal{M}\models\phi([g_1]_{\mathscr{U}},\dots,[g_n]_{\mathscr{U}})$ if and only if  $\big\{i\in I : \mathcal{M}_i\models \phi(g_1(i),\dots,g_n(i))  \big\}\in \mathscr{U}$.
Question. Does Łoś's theorem hold if the  ultrafilter $\mathscr{U}$ was a filter over $I$? Why?
 A: If you copy the ultraproduct construction, but using a filter instead of an ultrafilter, you get what's called a reduced product. As noted in the comments, Łoś's theorem does not hold for reduced products - there are problems with negation and disjunction. However, there are various weaker versions of Łoś's theorem which hold for reduced products. You can find information on this in section 6.2 of Model Theory by Chang and Keisler. A few years ago, I tried to summarize this information in an answer on this site.
The weaker versions of Łoś's theorem I referred to above all try to explain when $\prod_{i\in I}A_i/F\models \varphi(a)$ implies or is implied by containment of the set $\{i\in I\mid A_i\models \varphi(a_i)\}$ in the filter $F$. This is the most obvious way to try to generalize Łoś's theorem to reduced products. But the Feferman-Vaught theorem allows you to determine in a more subtle way when formulas hold in direct products (and reduced products more generally, as explained nicely in section 6.3 of Chang and Keisler).
Explicitly, the Feferman-Vaught theorem gives, for every formula $\varphi(x)$, a sequence of formulas $\theta_1(x),\dots,\theta_k(x)$ and a monotonic formula $\sigma(y_1,\dots,y_k)$ in the language of Boolean algebras, such that if $(A_i)_{i\in I}$ is an $I$-indexed family of structures and $F$ is a filter on $I$, and we set $||\theta_j(a)|| = \{i\in I\mid A_i\models \theta_j(a_i)\}\,/\,F \in \mathcal{P}(I)/F$, then $\prod_{i\in I}A_i/F\models \varphi(a)$ if and only if $\mathcal{P}(I)/F\models \sigma(||\theta_1(a)||,\dots,||\theta_k(a)||\}$. 
This statement can actually be viewed as a generalization of Łoś's theorem: In the case that $F$ is an ultrafilter, we can take just one formula $\theta_1(x) = \varphi(x)$ and set the formula $\sigma(y)$ to be $y = \top$. Then the theorem states that $\prod_{i\in I} A_i/F \models \varphi(a)$ if and only if $\mathcal{P}(I)/F \models ||\varphi(a)|| = \top$ if and only if $\{i\in I\mid A_i\models \varphi(a_i)\}\in F$, which is Łoś's theorem.
A: Look at the way in which maximality of the filter is used in the proof of Łoś's theorem. (BTW, the fact that it's spelled with Ł rather than with L is what tells you it's pronounced somewhat like the first syllable of "Washington".)
Suppose $\mathscr{U}$ is an ultrafilter.
If $\big\{i\in I : \mathcal{M}_i\models \phi(g_1(i),\dots,g_n(i))  \big\} \notin \mathscr{U}$ then $\big\{i\in I : \mathcal{M}_i \not\models \phi(g_1(i),\dots,g_n(i))  \big\} \in \mathscr{U}.$
In other words, if a set of indices fails to be a member of the ultrafilter, then its complement is a member. You can't do that if the filter is not maximal.
