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Let $S$ be the class of measurable simple functions $s$ on $\mathbb R^n$ such that $0\leq s(x)<\infty$ for all $x\in \mathbb R^n$.

Suppose $f:\mathbb R^n\to [0,\infty]$ is measurable. Then the integral of $f$ is $$\int fd\lambda=sup\left\{\int sd\lambda|s\leq s,s\in S\right\}$$

properties:

  1. $\int fd\lambda$ is well defined
  2. $0\leq \int fd\lambda \leq \infty$
  3. $\int cfd\lambda=c\int fd\lambda$
  4. if $f\leq g$, then $\int fd\lambda \leq \int gd\lambda$

Now I canwt prove $\int (f+g)d\lambda=\int fd\lambda+\int gd\lambda$

havakok
  • 1,067

2 Answers2

1

Well, we substantially have to choose one of the following paths: prove the claim directly by the definition (for example, we may want to show that both weak inequalities hold), which is a little tedious; or use a powerful convergence result (monotone convergence) plus the fact that the claim does hold for simple functions (i give the proof for $n=1$):

Let $f_n$, $g_n$ be two nondecreasing sequences of simple functions such that $\lim_{n} f_n(x)=f(x) $ and $\lim_{n} g_n(x) = g(x) $ for all $x \in \mathbb{R}$ ($f_n(x)= \frac{k-1}{2^n} $ if $\frac{k-1}{2^n} \le f(x) < \frac{k}{2^n}$ for every $k= 1,2 \dots n2^n $ and $f_n(x)=n $, it's easy to check that $f_n$ converges to $f$ ) . By monotone convergence and keeping in mind that $f_n$ and $g_n$ are simple for every $n$: $\int (f+g) = \int \lim (f_n+g_n) = \lim \int (f_n+g_n) = \lim \int f_n + \lim \int g_n = \int f + \int g $.

GaC
  • 2,498
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Clearly $\int(f+g) d\lambda \le \int f d\lambda + \int g d\lambda$.

For the other inequality. Given $\epsilon > 0$ there are $s_1 \le f$ and $s_2 \le g$ simple measurable functions such that $$\int s_1 d\lambda \ge \int f d\lambda - \frac{\epsilon}{2}$$ and $$\int s_2 d\lambda \ge \int g d\lambda - \frac{\epsilon}{2}.$$

Then $$\int f d\lambda + \int g d\lambda - \epsilon \le \int s_1 d\lambda + \int s_2 d\lambda = \int (s_1 + s_2) d\lambda \le \int(f+g)d\lambda,$$

where the last inequality follows from $s_1+s_2 \le f+g$ and the definition of the Lebesgue integral. Overall

$$ \int f d\lambda + \int g d\lambda \le \int(f+g) d\lambda + \epsilon.$$

Since this holds for all $\epsilon >0,$ we obtain the required inequality . Hope this helps.

Zestylemonzi
  • 4,103