# Self Study Measure Theory: Proof Not An Algebra

I am struggling with the following statement:

Proof that:

$$\mathcal{J} = \{(a,b] \subseteq (0, 1] : a,b \in \mathbb{R}\}$$ is not an algebra (or field).

$$\mathcal{J}$$ is the set of all half open intervals in $$(a,b] \in (0,1]$$. To be an algebra the set has to fulfill the following condition:

1. $$\emptyset \in \mathcal{J}$$ which is true, since $$(0, 1]$$ is also in $$\mathcal{J}$$ and since the algebra is closed under set difference then $$(0, 1] \setminus (0,1] = \emptyset$$
2. $$A_i = (a_i, b_i], A_j = (a_j, b_j], \text{ wlog let } A_i \cap A_j \neq \emptyset$$ then wlog let $$0 < a_i, a_j < b_i \leq b_j \leq 1$$, then $$A_i \cup A_j = (\min\{a_i, a_j\}, \max\{b_i, b_j\}]$$ which is a half open interval of the same kind. Thus the set is closed under pairwise union.

Thus there remains only point three of the definition of an algebra, which is closed under difference. So this has to fail if I am not mistaken thus far. However this should be true as well:

1. Let $$A_i = (a_i, b_i], A_j = (a_j, b_j], \ A_i \cap A_j = \emptyset$$ then $$A_j \setminus A_i = A_j \in \mathcal{J}$$ and if $$A_i \cap A_j \neq \emptyset$$ then wlog let $$0 < a_i, a_j < b_i \leq b_j \leq 1$$ and $$A_j \setminus A_i = (b_i, b_j]$$ which is again an half open interval.

Concluding, either way there ends up to be an half open interval. So this should be an Algebra, but it isn't. So there must be a mistake in one of my arguments.

• $(0,1/3]$ and $(2/3,1]$ both belong to $\mathcal J$ but their union does not. Commented Dec 28, 2021 at 12:12
• because it's not an interval? Commented Dec 28, 2021 at 13:13
• Yes. Members of J are intervals. Commented Dec 29, 2021 at 8:28
• And i was only thinking in terms of sets. i guess i didn't see the forest for trees. thx. Commented Dec 29, 2021 at 10:33

$$\emptyset \in \mathcal{J}$$ which is true, since $$(0, 1]$$ is also in $$\mathcal{J}$$

Is it? Why would $$(0,1]\in \mathcal J$$ imply that $$\emptyset\in\mathcal J$$? You state this as if the implication is clear, but I don't see why this is the case.

$$A_i = (a_i, b_i], A_j = (a_j, b_j], \text{ wlog let } 0 < a_i,a_j < b_i,b_j \leq 1$$, then $$A_i \cup A_j = (\min\{a_i, a_j\}, \max\{b_i, b_j\}]$$

Here, again, you present a claim without proving it. In particular, you claim that two sets, $$A_i \cup A_j$$ and $$(\min\{a_i, a_j\}, \max\{b_i, b_j\}]$$, are identical, but you do not prove that they are.

In fact, they are not always identical. It is always true that $$A_i \cup A_j\subseteq (\min\{a_i, a_j\}, \max\{b_i, b_j\}]$$, but the converse is not always true.

Let $$A_i = (a_i, b_i], A_j = (a_j, b_j], \ A_i \cap A_j = \emptyset$$ then $$A_j \setminus A_i = A_j \in \mathcal{J}$$ and if $$A_i \cap A_j \neq \emptyset$$ then wlog let $$0 < a_i, a_j < b_i \leq b_j \leq 1$$ and $$A_j \setminus A_i = (b_i, b_j]$$ which is again an half open interval.

Again, you present a claim and no proof. And as above, some of the claims here are just plain wrong. If you try to prove each of the equalities you claim, you should be able to identify which are not true.

• can you give me a hint? i really don't get my mistake. Commented Dec 28, 2021 at 10:48
• @user2550228 The hint is that you should try to prove the equalities you claim. You should quite soon run into problems that will allow you to find counterexmaples.
– 5xum
Commented Dec 28, 2021 at 10:56