Let us agree that for events A_i,A_j,A_k property b_(ijk)…

Let us agree that for events $A_i,A_j,A_k$ property $b_{ijk}$ holds for $A_i, A_j, A_k$ if \[ 1^{\circ}\ \ A_i\,\overline{A_j}\,A_k = V, \qquad 2^{\circ}\ \ \overline{A_i}\,A_j\,\overline{A_k} = V, \] where $V$ denotes the impossible event. Prove:
1) if property $b_{ijk}$ holds, then $\varrho(A_i,A_j)+\varrho(A_j,A_k)=\varrho(A_i,A_k)$, where $\varrho(A,B)=P(A\triangle B)$ (Problem 63);
2) if $P(A)=1$ implies $A=E$ (the certain event), then the converse also holds.

Step 1: The definition and its meaning

About the definition. In the text as received, the conditions read "$A_i\cap\overline{A_j\cap A_k}=V$" and "$\overline{A_i}\cap\overline{A_j}\cap\overline{A_k}=V$". With those conditions the statements of Problems 64–68 are false (for example $A_i=V$, $A_j=X$, $A_k=\overline{X}$ with $P(X)=\tfrac12$ satisfies them, yet $\varrho(A_i,A_j)+\varrho(A_j,A_k)=\tfrac32\ne\tfrac12=\varrho(A_i,A_k)$). The bars were evidently misplaced; the definition above is the one for which all these problems are true. It says: $A_j$ lies "between" $A_i$ and $A_k$, i.e. $A_iA_k\subseteq A_j\subseteq A_i+A_k$.

In words: there is no outcome where $A_i$ and $A_k$ both occur but $A_j$ does not, and no outcome where $A_j$ occurs but neither $A_i$ nor $A_k$ does. So $A_j$ contains the common part of $A_i$ and $A_k$ and lies inside their sum.

Step 2: The key identity (Problem 63)

Splitting $\Omega$ into the eight atoms of $A_i,A_j,A_k$ gives, for any events,

\[ \varrho(A_i,A_j)+\varrho(A_j,A_k)-\varrho(A_i,A_k) = 2\Big[P\big(A_i\overline{A_j}A_k\big)+P\big(\overline{A_i}A_j\overline{A_k}\big)\Big]. \tag{$*$} \]

(On each atom, "disagree with $A_j$" is counted on the left and "disagree with each other" on the right; they differ only when $A_i$ and $A_k$ agree and $A_j$ disagrees with both.)

Step 3: Part 1 — $b_{ijk}$ implies equality

If $b_{ijk}$ holds, both atoms in $(*)$ are the impossible event, so their probabilities are $0$ and

\[ \boxed{\varrho(A_i,A_j)+\varrho(A_j,A_k)=\varrho(A_i,A_k).} \]

Step 4: Part 2 — the converse

Suppose the equality holds. By $(*)$,

\[ P\big(A_i\overline{A_j}A_k\big)+P\big(\overline{A_i}A_j\overline{A_k}\big) = 0, \]

and since probabilities are non-negative, each is $0$. Now use the assumption "$P(A)=1\Rightarrow A=E$". Applied to the opposite event $\overline{X}$ it reads: $P(X)=0\Rightarrow X=V$. Hence

\[ A_i\overline{A_j}A_k = V, \qquad \overline{A_i}A_j\overline{A_k}=V, \]

which is property $b_{ijk}$. $\blacksquare$

Step 5: Why the extra assumption is needed

Without it the converse can fail. Choose a point uniformly in $[0,1]$; let $A_i=A_k=V$ and $A_j=\{\tfrac12\}$. All distances are $0$, so the equality holds, but $\overline{A_i}A_j\overline{A_k}=\{\tfrac12\}\ne V$: property $b_{ijk}$ fails on an event of probability $0$.

Step 6: Example

Die: $A_i=\{1,2,3\}$, $A_j=\{2,3,4\}$, $A_k=\{3,4,5\}$. $A_i\overline{A_j}A_k = \{1,2,3\}\cap\{1,5,6\}\cap\{3,4,5\}=\varnothing$ and $\overline{A_i}A_j\overline{A_k}=\{4,5,6\}\cap\{2,3,4\}\cap\{1,2,6\}=\varnothing$. So $b_{ijk}$ holds, and indeed $\tfrac26+\tfrac26=\tfrac46$ (Problem 63, Step 6).