Kind of passage to the limit in the sense of distributions

Question

Suppose $B$ is a ball in $\mathbb{R}^{n}$ with $n>1$, and $f$ a locally integrable function. Suppose $F$ is a closed set with empty interior and $M>0$ such that the distribution defined by $f$ satisfies \begin{equation} \int_{B\setminus F}f(t)\phi(t)dt\leq M \end{equation} for all test function $\phi$. We know that if we had $$f(x)\leq M$$ for all $x\in B\setminus F$, we could get the same inequality by passing to the limit (supposing that $f$ is continuous); is it possible, in the same way, to circumvent around $F$ in (1) and prove that (1) holds for integrals over $B$?