regarding lebesgue integral

Question

as in the wikipedia page, we define the Lebesgue integralon non-negative functions: Let f be a non-negative measurable function on E, which we allow to attain the value +∞, in other words, $f$ takes non-negative values in the extended real number line. We define: $\int_E f d\mu= sup{\int_E s d\mu :0 \leq s\leq f , s simple}$.

I want to prove using this definition that:

1. The lebesgue integral of an indicator function of a measurable set $A$ is $m(A)$ : $\int_A 1 dm= m(A)$.

2. $|\int f dm| \leq \int |f|dm$.

3. If $A, B$ are disjoint sets then $\int_ {A\cup B} fdm= \int_A fdm+\int_B fdm$.

4. $\int (af+bg)dm=a\int f dm + b\int g dm$ where $a,b\in R$.

I am wondering if it is possible to show all these properties using only the definition of lebesgue integral I wrote (with sup)?

In 4: it is sufficient to show that: $\int cf=c\int f$ for $c \geq 0$ and that $\int (f+g)=\int f+\int g$ for $f,g$ measurable positive functions. So, $\int cf=sup_{0\leq s\leq cf} \int s dm=sup_{0\leq 1/c*s\leq f} c \int 1/c *s dm=c sup_{0\leq 1/c*s\leq f} \int 1/c*s= c \int fdm$. And if $s_1\leq f$ and $s_2\leq g$ then $s_1+s_2\leq f+g$ so $\int f+\int g \leq \int (f+g)$, what about the other side?

For 2: We have: $\int |f|dm = sup{\int s: 0 \leq s\leq |f|, s simple}$ and $|\int f dm|=|sup {\int s : 0\leq s \leq f, s simple}|$.

Now, we know that if $s\leq f$ then $s\leq |f|$ since $f\leq |f|$, thus $|\int f dm| \leq \int |f|dm$. Is is fine?

For 3, we know that $A, B \subset A\cup B$ thus $\int_A f, \int_B f\leq \int_{A\cup B} f$ (I think that it is obvious by the def of the lebesgje integral, however i am not sure of how to show it formally) therefore $\int_A f+\int_B f \leq \int_{A\cup B} f$. What about the other side?

Thank you very much in advance.

Answer 1

Let $(\mathrm{X}, \mathscr{X}, \mu)$ be a measure space and suppose we want to define the integral for the first time. To do so, we start with your 1. as definition of integral of an indicator function.

1. By definition, if $\mathrm{A} \in \mathscr{X},$ then $\int\limits_\mathrm{A} 1 d\mu = \mu(\mathrm{A}).$
2. Since $1_{\mathrm{A} \cup \mathrm{B}} = 1_\mathrm{A} + 1_\mathrm{B}$ for disjoint $\mathrm{A}, \mathrm{B} \in \mathscr{X},$ and since the integral is aimed at being a linear form, we define the integral of a simple function $\sum c_k 1_{\mathrm{A}_k},$ where the $\mathrm{A}_k$ are pairwise-disjoint, as $\int \sum c_k 1_{\mathrm{A}_k} d\mu = \sum c_k \int\limits_\mathrm{A_k} 1 d\mu = \sum c_k \mu(\mathrm{A}_k).$
3. If $f \geq 0,$ define $\int f d\mu$ according to what you wrote, in other words, $\int fd\mu = \sup\limits_{0 \leq s \leq f} \int s d\mu$ where $s$ is a simple function. We also define, $\int\limits_\mathrm{A} fd\mu = \int 1_\mathrm{A} f d\mu.$

To your properties:

• As already mentioned: 1. is definition.

• As for number 2., note that by definition, if $s \geq 0$ is simple, then $\int s d\mu \geq 0$ and therefore, $\int f d\mu \geq 0$ when $f \geq 0,$ so your 2. follows.

• To prove your number 3, note, as mentioned above, $1_{\mathrm{A} \cup \mathrm{B}} = 1_\mathrm{A} + 1_\mathrm{B},$ so if you multiply both sides by a simple function $0 \leq s \leq f,$ it also breaks as a sum and this sum is therefore simple as well. Then, the definition os integral of simple function kicks in and allows you to write the integral as follows $\int 1_{\mathrm{A} \cup \mathrm{B}} s d\mu = \int 1_\mathrm{A} s d\mu + \int 1_\mathrm{B}s d\mu.$ Also, you need to realise that if $0 \leq s \leq f$ is simple then $\mathrm{1}_A s$ and $1_\mathrm{B} s$ are also simple and if $s_A$ and $s_B$ satisfy that $0 \leq s_A \leq 1_\mathrm{A} f$ and $0 \leq s_B \leq 1_\mathrm{B} f$ are simple function then $s = s_A + s_B$ satisfies $0 \leq s \leq f$ is also simple. The result follows. (You probably need to recall elementary 1st year calculus results $\sup (\mathrm{S} + \mathrm{T}) = \sup \mathrm{S} + \sup \mathrm{T}$ when $\mathrm{S}$ and $\mathrm{T}$ are sets of positive real numbers.)

• To prove linearity you need the monotone convergence theorem.That is why you were probably stuck. In that case, linearity follows by linearty in simple functions and then a monotone convergence theorem.