# 微积分网课代修|积分学代写Integral Calculus代考|MATH122 Null or negligible sets

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## 微积分作业代写calclulus代考|Null or negligible sets

We define a null set to be those sets on which the simplest function $F(x)=x$ does not grow. 3 This can be given its own definition. If $F(x)$ then
$$\left|F\left(b_{i}\right)-F\left(a_{i}\right)\right|=\left(b_{i}-a_{i}\right)$$
so that the failure of $F$ to grow on a set $E$ can be described simply by using the sums
$$\sum_{i=1}^{n}\left(b_{i}-a_{i}\right)$$
taken over a subpartition.
DEFINITION 3.4. A set $E$ is said to be a negligible set (or a null set) if for every $\epsilon>0$ there can be found a full cover $\beta$ of that set $E$ so that
$$\sum_{i=1}^{n}\left(b_{i}-a_{i}\right)<\epsilon$$
whenever the collection
$$\gamma=\left{\left(\left[a_{i}, b_{i}\right], x_{i}\right): i=1,2, \ldots, n\right}$$
is a subpartition chosen from $\beta$.

## 微积分作业代写calclulus代考|Exercises.

EXERCISE 25 . Show that a set $E$ is a null set if and only if the function $F(x)=x$ does not grow on $E$.
SOLUTION IN SECTION 8.3.4.
EXERCISE 26. Show that a Lipschitz function cannot grow on a null set.
SOLUTION IN SECTION 8.3.7.
EXERCISE 27. Suppose that $F: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable at each point of a set $E$. Show that then $F$ cannot grow on any null subset of $E$.
SOLUTION IN SECTION 8.3.11.
EXERCISE 28. Show that every subset of a null set is a null set.

## 微积分作业代写calclulus代考| Null or negligible sets

$$\left|F\left(b_{i}\right)-F\left(a_{i}\right)\right|=\left(b_{i}-a_{i}\right)$$

$$\sum_{i=1}^{n}\left(b_{i}-a_{i}\right)$$

$$\sum_{i=1}^{n}\left(b_{i}-a_{i}\right)<\epsilon$$