This is exercise from Megginson's Banach Space Theory.
I have to show that if $B$ is a Borel subset of a topological space $X$, then there exist sets $M, N$ of first category in $X$ such that $(B\setminus M)\cup N$ is open. As far as I know a Borel sigma algebra is the smallest sigma algebra containing all open subsets of a topological space and every member of the Borel sigma algebra is called a Borel set. But can a Borel set be represented by sets of first category and open sets? I have no idea. Please help.
Consider the set
$$\mathcal{F} = \{A\subseteq X : \exists_{M,N} \text{ of first category such that } A\backslash M \cup N \text{ is open. } \}$$
It contains $X$,$\emptyset$. It is closed under countable unions (simply because countable unions of sets of the first category are of the first category, By Baire's theorem). Finally it is closed under complements (I prove below). Hence $\mathcal{F}$ is a $\sigma$-algebra.
Moreover $\mathcal{F}$ trivially contains all open subsets and so all Borel subsets.
Why is it closed under complements: This is quite messy.
Let $A\in\mathcal{F}$ so there exists $M,N$ such that $A\backslash M \cup N$ is open. Hence $X\backslash (A\backslash M \cup N)= (X\backslash A\cup M)\backslash N = (X\backslash A) \backslash N \cup M\backslash N$ is closed, $N,M\backslash N$ are of the first category.
Denote this closed set by $C$. Now let $L:= C\backslash \text{int}(C)$ so $L$ is of first category and $C\backslash L$ is open. So closed sets are in $\mathcal{F}$. Therefore $(X\backslash A)\backslash N \cup M\backslash N$ is in $\mathcal{F}$. Since $N,M\backslash N$ are of first category we also have that $X\backslash A$ in $\mathcal{F}$ as desired.