 Solution

Boundedness of $f$ makes $\phi$ finite and nonnegative. Suppose $\phi (x_0)<a$. By its infimum definition, some $n$ satisfies ${\phi }_n(x_0)<a$. If $|x-x_0|<1/(2n)$, thenThe supremum over the smaller interval cannot exceed that over the larger one. HenceThus $\{\phi <a\}$ is open for every $a$, proving $\phi$ is upper semicontinuous. Measurability of the original $f$ was never needed: these are purely pointwise oscillation bounds.