Understanding continuity lemma

Question

In Mesure Theory, we have a continuity lemma that says $$\text{if} \ u\colon (a,b) \times X \rightarrow \mathbb{R} \ \text{ satisfies that} \ x \mapsto u(t,x) \in \mathcal{L}^1(\mu) \forall t, \ t \mapsto u(t,x) \ \text{is continuous for all }x \in X, \ \text{and} |u(t,x)| \le w(x) \ \forall (t,x) \in (a,b) \ \text{and some} \ w \in \mathcal{L}^1_+(\mu), $$ then the following function is continuous $$t \mapsto \int u(t,x) \ \mu(dx)$$ Okay, but what if $u$ is just a regular function that is not multivariate, and depends on $x$ alone? If $u$ is Lebesgue integrable, can we then write $u(x) = u(t,x)$, and conclude that $t \mapsto \int u(x) \mu(dx)$ is continuous?