I'll assume that $f$ takes values in $\mathbb{R}$, and leave it to you to adjust in the case where $f$ actually takes values in the extended reals.
If $a\geq 0$, then $\chi_Bf(x)\gt a$ if and only if $f(x)\gt a$ and $\chi_B(x)=1$; that is, $[\chi_Bf\gt a] = [f\gt a]\cap B\text{ if }a\geq 0.$ Since $f$ is measurable, and $B$ is measurable, the intersection is measurable.
If $a\lt 0$, then $\chi_B f(x)\gt a$ if and only if $x\in B$ and $f(x)\gt a$, or $\chi_B(x)=0$ (since then $\chi_Bf(x)=0\gt a$). So $[\chi_Bf\gt a] = ([f\gt a]\cap B) \cup (\mathbb{R}-B)\text{ if }a\lt 0.$ This is a union of two measurable sets, hence measurable.