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Two densely defined traces on a $C^*$-algebra coinciding on a dense subalgebra are equal

Let $t_1$ and $t_2$ be lower semicontinuous semifinite densely defined traces on a $C^*$-algebra $A$. Let us denote by $\mathcal{R}_1$ and $\mathcal{R}_2$ their ideals of definition, i.e. $\mathcal{R}_i=\left\{x\in A\ \middle|\ t_i(x^*x)<+\infty \right\}$. Next, let $B$ be an involutive dense subalgebra of $A$ such that $B\subset \mathcal{R}_1,\mathcal{R}_2$ and $t_1(b)=t_2(b)$ for any $b\in B^+$. Is it true that then $t_1=t_2$ on the whole algebra $A$?

I have found a similar statement in “C*-algebras” by J.Dixmier 1977: Lemma 6.5.3 on page 139. But he formulates this claim in terms of bitraces, which look artificial to me. Can anyone provide a straightforward and concise argument without bitraces? Is there a better reference for this?