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(1) Existence of sets of small measure.

Let $(\Omega,\Sigma,\mu)$ an atomless finite measure space with $\mu(\Omega)>0$. It is easy to show that any measurable set $A$ with $\mu(A)>0$ contains a measurable subset $B\subset A$ with $0<\mu(B)\le \mu(A)/2$, and therefore a measurable set $M$ of measure $0<\mu(M)\le \varepsilon$.

(2) The space and any measurable set is a numerable union of set of small measure.

Now given $\varepsilon>0$ consider the set $\mathcal M$ of measurable sets $M=\bigcup_n A_n$ union of a numerable set of measurable sets of measure $\mu(A_n)\le \varepsilon$

Let $\lambda$ the supremum of the measures $\mu(M)$ of set in $\mathcal M$. There is a sequence $(M_n)$ of sets in $\mathcal M$ with $\lim_n\mu(M_n)=\lambda$. Then $M_0:=\bigcup M_n\in \mathcal M$ (by definition) and $\lambda(M_0)=\lambda$. But it follows that $\lambda=\mu(\Omega)$, because in other case $\Omega\smallsetminus M_0$ will contain a set of measure $\le \varepsilon$, which contradict the maximality of $M_0$.

(3) Construct a set of measure $a$.

Now given $0<a<\mu(\Omega)$ we construct an increasing sequence $(A_n)$ of measurable sets with $0<a-\mu(A_n)<\varepsilon_n$ for any sequence of strictly decreasing positive numbers with $\varepsilon_n\to0$ and $\varepsilon_1<a$

Since $\Omega=\bigcup_n M_n$ is a union of measurable sets of measure $\le \varepsilon_1$, we may take $A_1=\bigcup_{n=1}^N M_n$, where $N$ is the greatest integer such that the measure of the union is less than or equal $a$. To construct $A_2$ consider $\Omega\smallsetminus A_1$ and construct $B_2\subset \Omega\smallsetminus A_1$ with $(a-\mu(A_1))-\mu(B_2)\le\varepsilon_2-\varepsilon_1$ and put $A_2=A_1\cup B_2$. The construction follows in the same way from this point.

Notice that the construction of $B_2$ and that of $A_1$ are totally similar.

(1) Existence of sets of small measure.

Let $(\Omega,\Sigma,\mu)$ an atomless finite measure space with $\mu(\Omega)>0$. It is easy to show that any measurable set $A$ with $\mu(A)>0$ contains a measurable subset $B\subset A$ with $0<\mu(B)\le \mu(A)/2$, and therefore a measurable set $M$ of measure $0<\mu(M)\le \varepsilon$.

(2) The space and any measurable set is a numerable union of set of small measure.

Now given $\varepsilon>0$ consider the set $\mathcal M$ of measurable sets $M=\bigcup_n A_n$ union of a numerable set of measurable sets of measure $\mu(A_n)\le \varepsilon$

Let $\lambda$ the supremum of the measures $\mu(M)$ of set in $\mathcal M$. There is a sequence $(M_n)$ of sets in $\mathcal M$ with $\lim_n\mu(M_n)=\lambda$. Then $M_0:=\bigcup M_n\in \mathcal M$ (by definition) and $\lambda(M_0)=\lambda$. But it follows that $\lambda=\mu(\Omega)$, because in other case $\Omega\smallsetminus M_0$ will contain a set of measure $\le \varepsilon$, which contradict the maximality of $M_0$.

(3) Given a measurable set $M$ and two numbers $0<a<b<\mu(M)$, there exists a measurable set $A\subset M$ with $a<\mu(A)<b$.

Let $\varepsilon=(b-a)/2$, by (2) the set $M$ can be put as a union of measurable sets of measure less than $\varepsilon$, $M=\bigcup_n M_n$ with $\mu(M_n)\le \varepsilon$. Then $$\lim_N\mu\Bigl(\bigcup_{n=1}^N M_n\Bigr)=\mu(M).$$ Therefore there is some $N$ such that $a<\mu\Bigl(\bigcup_{n=1}^N M_n\Bigr)<b$, since each $M_n$ have measure $\le (b-a)/2$.

(4) Construct the set with measure $\mu(A)=a$ for a given $0<a<\mu(\Omega)$.

We will construct a sequence of measurable sets $(A_n)$ with $A_n\le A_{n+1}$ and such that $$\frac{a+\mu(A_n)}{2}<\mu(A_{n+1})< a.$$ It is clear that in this case $A=\bigcup_n A_n$ have measure $a$.

Start with $A_0=\emptyset$. We must construct $A_1$ with $a/2<\mu(A_1)<a$. Sincer $\Omega$ have measure $>a$, we have constructed such a set in (3).

Assume we have constructed $A_n$, then $\mu(A_n)<a$, applying (3) we construct $B_n\subset \Omega\smallsetminus A_n$ whose measure is greater than half the difference with our objective: $a-\mu(A_n)>\mu(B_n)>\frac{a-\mu(A_n)}{2}$. Hence $A_{n+1}=A_n\cup B_n$ satisfies our requirement. First $A_n$ and $B_n$ are disjoint $\mu(A_{n+1})=\mu(A_n)+\mu(B_n)<a$, and $\mu(A_n)+\mu(B_n)>\frac{a+\mu(A_n)}{2}$.

(1) Existence of sets of small measure.

Let $(\Omega,\Sigma,\mu)$ an atomless finite measure space with $\mu(\Omega)>0$. It is easy to show that any measurable set $A$ with $\mu(A)>0$ contains a measurable subset $B\subset A$ with $0<\mu(B)\le \mu(A)/2$, and therefore a measurable set $M$ of measure $0<\mu(M)\le \varepsilon$.

(2) The space and any measurable set is a numerable union of set of small measure.

Now given $\varepsilon>0$ consider the set $\mathcal M$ of measurable sets $M=\bigcup_n A_n$ union of a numerable set of measurable sets of measure $\mu(A_n)\le \varepsilon$

Let $\lambda$ the supremum of the measures $\mu(M)$ of set in $\mathcal M$. There is a sequence $(M_n)$ of sets in $\mathcal M$ with $\lim_n\mu(M_n)=\lambda$. Then $M_0:=\bigcup M_n\in \mathcal M$ (by definition) and $\lambda(M_0)=\lambda$. But it follows that $\lambda=\mu(\Omega)$, because in other case $\Omega\smallsetminus M_0$ will contain a set of measure $\le \varepsilon$, which contradict the maximality of $M_0$.

(3) Construct a set of measure $a$.

Now given $0<a<\mu(\Omega)$ we construct an increasing sequence $(A_n)$ of measurable sets with $0<a-\mu(A_n)<\varepsilon_n$ for any sequence of strictly decreasing positive numbers with $\varepsilon_n\to0$ and $\varepsilon_1<a$

Since $\Omega=\bigcup_n M_n$ is a union of measurable sets of measure $\le \varepsilon_1$, we may take $A_1=A\cup\bigcup_{n=1}^N M_n$, where $N$ is the last number such that the measure of the union is less than or equal $a$. To construct $A_2$ consider $\Omega\smallsetminus A_1$ and construct $B_2\subset \Omega\smallsetminus A_1$ with $(a-\mu(A_1))-\mu(B_2)\le\varepsilon_2-\varepsilon_1$ and put $A_2=A_1\cup B_2$. The construction follows in the same way from this point.

Notice that the construction of $B_2$ and that of $A_1$ are totally similar.

(1) Existence of sets of small measure.

Let $(\Omega,\Sigma,\mu)$ an atomless finite measure space with $\mu(\Omega)>0$. It is easy to show that any measurable set $A$ with $\mu(A)>0$ contains a measurable subset $B\subset A$ with $0<\mu(B)\le \mu(A)/2$, and therefore a measurable set $M$ of measure $0<\mu(M)\le \varepsilon$.

(2) The space and any measurable set is a numerable union of set of small measure.

Now given $\varepsilon>0$ consider the set $\mathcal M$ of measurable sets $M=\bigcup_n A_n$ union of a numerable set of measurable sets of measure $\mu(A_n)\le \varepsilon$

Let $\lambda$ the supremum of the measures $\mu(M)$ of set in $\mathcal M$. There is a sequence $(M_n)$ of sets in $\mathcal M$ with $\lim_n\mu(M_n)=\lambda$. Then $M_0:=\bigcup M_n\in \mathcal M$ (by definition) and $\lambda(M_0)=\lambda$. But it follows that $\lambda=\mu(\Omega)$, because in other case $\Omega\smallsetminus M_0$ will contain a set of measure $\le \varepsilon$, which contradict the maximality of $M_0$.

(3) Construct a set of measure $a$.

Now given $0<a<\mu(\Omega)$ we construct an increasing sequence $(A_n)$ of measurable sets with $0<a-\mu(A_n)<\varepsilon_n$ for any sequence of strictly decreasing positive numbers with $\varepsilon_n\to0$ and $\varepsilon_1<a$

Since $\Omega=\bigcup_n M_n$ is a union of measurable sets of measure $\le \varepsilon_1$, we may take $A_1=\bigcup_{n=1}^N M_n$, where $N$ is the greatest integer such that the measure of the union is less than or equal $a$. To construct $A_2$ consider $\Omega\smallsetminus A_1$ and construct $B_2\subset \Omega\smallsetminus A_1$ with $(a-\mu(A_1))-\mu(B_2)\le\varepsilon_2-\varepsilon_1$ and put $A_2=A_1\cup B_2$. The construction follows in the same way from this point.

Notice that the construction of $B_2$ and that of $A_1$ are totally similar.

This is how we solved this didactical problem in the University of Sevilla.

(1) Existence of sets of small measure.

Let $(\Omega,\Sigma,\mu)$ an atomless finite measure space with $\mu(\Omega)>0$. It is easy to show that any measurable set $A$ with $\mu(A)>0$ contains a measurable subset $B\subset A$ with $0<\mu(B)\le \mu(A)/2$, and therefore a measurable set $M$ of measure $0<\mu(M)\le \varepsilon$.

(2) The space and any measurable set is a numerable union of set of small measure.

Now given $\varepsilon>0$ consider the set $\mathcal M$ of measurable sets $M=\bigcup_n A_n$ union of a numerable set of measurable sets of measure $\mu(A_n)\le \varepsilon$

Let $\lambda$ the supremum of the measures $\mu(M)$ of set in $\mathcal M$. There is a sequence $(M_n)$ of sets in $\mathcal M$ with $\lim_n\mu(M_n)=\lambda$. Then $M_0:=\bigcup M_n\in \mathcal M$ (by definition) and $\lambda(M_0)=\lambda$. But it follows that $\lambda=\mu(\Omega)$, because in other case $\Omega\smallsetminus M_0$ will contain a set of measure $\le \varepsilon$, which contradict the maximality of $M_0$.

(3) Construct a set of measure $a$.

Now given $0<a<\mu(\Omega)$ we construct an increasing sequence $(A_n)$ of measurable sets with $0<a-\mu(A_n)<\varepsilon_n$ for any sequence of strictly decreasing positive numbers with $\varepsilon_n\to0$ and $\varepsilon_1<a$

Since $\Omega=\bigcup_n M_n$ is a union of measurable sets of measure $\le \varepsilon_1$, we may take $A_1=A\cup\bigcup_{n=1}^N M_n$, where $N$ is the last number such that the measure of the union is less than or equal $a$. To construct $A_2$ consider $\Omega\smallsetminus A_1$ and construct $B_2\subset \Omega\smallsetminus A_1$ with $(a-\mu(A_1))-\mu(B_2)\le\varepsilon_2-\varepsilon_1$ and put $A_2=A_1\cup B_2$. The construction follows in the same way from this point.

Notice that the construction of $B_2$ and that of $A_1$ are totally similar.

(1) Existence of sets of small measure.

Let $(\Omega,\Sigma,\mu)$ an atomless finite measure space with $\mu(\Omega)>0$. It is easy to show that any measurable set $A$ with $\mu(A)>0$ contains a measurable subset $B\subset A$ with $0<\mu(B)\le \mu(A)/2$, and therefore a measurable set $M$ of measure $0<\mu(M)\le \varepsilon$.

(2) The space and any measurable set is a numerable union of set of small measure.

Now given $\varepsilon>0$ consider the set $\mathcal M$ of measurable sets $M=\bigcup_n A_n$ union of a numerable set of measurable sets of measure $\mu(A_n)\le \varepsilon$

Let $\lambda$ the supremum of the measures $\mu(M)$ of set in $\mathcal M$. There is a sequence $(M_n)$ of sets in $\mathcal M$ with $\lim_n\mu(M_n)=\lambda$. Then $M_0:=\bigcup M_n\in \mathcal M$ (by definition) and $\lambda(M_0)=\lambda$. But it follows that $\lambda=\mu(\Omega)$, because in other case $\Omega\smallsetminus M_0$ will contain a set of measure $\le \varepsilon$, which contradict the maximality of $M_0$.

(3) Construct a set of measure $a$.

Now given $0<a<\mu(\Omega)$ we construct an increasing sequence $(A_n)$ of measurable sets with $0<a-\mu(A_n)<\varepsilon_n$ for any sequence of strictly decreasing positive numbers with $\varepsilon_n\to0$ and $\varepsilon_1<a$

Since $\Omega=\bigcup_n M_n$ is a union of measurable sets of measure $\le \varepsilon_1$, we may take $A_1=A\cup\bigcup_{n=1}^N M_n$, where $N$ is the last number such that the measure of the union is less than or equal $a$. To construct $A_2$ consider $\Omega\smallsetminus A_1$ and construct $B_2\subset \Omega\smallsetminus A_1$ with $(a-\mu(A_1))-\mu(B_2)\le\varepsilon_2-\varepsilon_1$ and put $A_2=A_1\cup B_2$. The construction follows in the same way from this point.

Notice that the construction of $B_2$ and that of $A_1$ are totally similar.