**Here is a test:**

We will divide our interval into 2 intervals. $ (- 1, 0) $ and $ (0.1) $:

(1) for the interval $ (- 1, 0) $, our function is increasing in $ (- 1.0) $ then it is differentiable a.e. in $ (- 1.0) $ for Lebesgue's theorem on p.112 and its derivative is $ f & # 39; (x) = frac {1} {3} x ^ {2/3} $ and

our function $ f $ it's continuous

(2) for the interval $ (0, 1) $.

We will do this with the help of problem 37 on p. 123. Let $ epsilon> 0. $ taking $ {(c_ {i}, d_ {i}): 1 leq i leq n } $ be a collection of non-overlapping intervals in $ (0.1) $ such that $ sum_ {i = 1} ^ {n} (d_ {i} – c_ {i}) < epsilon ^ 3. $ Choose $ a = frac { epsilon ^ 3} {8}. $ Now we break the sum $ sum_ {i = 1} ^ {n} | f (d_ {i}) – f (c_ {i}) | $ in two parts, those intervals that are in $ (0, a) $ and those in $ (a, 1). $ that is, we break the interval in $ to $ Similar to $ eqn. (19) $ on page 117. Let $ a = d_ {m} $ for some $ m. $

Now consider the sum over the intervals that are in $ (0, a), $ $$ sum_ {i = 1} ^ {m} | f (d_ {i}) – f (c_ {i}) | = sum_ {i = 1} ^ {m} | d_ {i} ^ {1/3} – c_ {i} ^ {1/3} | leq a ^ {1/3} = epsilon / 2. $$

This follows from the fact that $ x ^ 1/3 $ It is a growing function. Monotonicity ensures that the function does not oscillate widely.

Now we consider the sum of the intervals that are in $ (a, 1). $

$$ sum_ {i = m + 1} ^ {n} | f (d_ {i}) – f (c_ {i}) | = sum_ {i = m + 1} ^ {n} | d_ {i} ^ {1/3} – c_ {i} ^ {1/3} | = sum_ {i = m + 1} ^ {n} | d_ {i} ^ {1/3} – c_ {i} ^ {1/3} | times frac {| (d_ {i} ^ {2/3} + (d_ {i} times c_ {i}) ^ {1/3} + c_ {i} ^ {2/3}) |} {| (d_ {i} ^ {2/3} + (d_ {i} times c_ {i}) ^ {1/3} + c_ {i} ^ {2/3}) |} $$

$$ = sum_ {i = m + 1} ^ {n} frac {(d_ {i} – c_ {i})} {| (d_ {i} ^ {2/3} + (d_ {i} times c_ {i}) ^ {1/3} + c_ {i} ^ {2/3}) |} $$

So,$ sum_ {i = m + 1} ^ {n} | f (d_ {i}) – f (c_ {i}) | leq sum_ {i = m + 1} ^ {n} frac {(d_ {i} – c_ {i})} {8 a ^ {2/3}} = ** frac {1} {8 a ^ 2/3} ** sum_ {i = m + 1} ^ {n} (d_ {i} – c_ {i}) < frac {1} {2 epsilon ^ 2}. epsilon ^ 3 = frac { epsilon} {2}. $

We use $ (xy) = (x 1/3 – y 1/3) (x 2/3 + (xy) 1/3 + y 2/3). $

Combining these two sums we see that

$ sum_ {i = 1} ^ {n} | f (d_ {i}) – f (c_ {i}) | leq ( sum_ {i = 1} ^ {m} | f (d_ {i}) – f (c_ {i}) | + sum_ {i = m + 1} ^ {n} | f (d_ { i}) – f (c_ {i}) |) < epsilon. $

**My question is:**

1-It turns out that this part of my judgment $ frac {1} {8 to 2/3} $ Is it wrong, could someone help me adjust it, please?

Note that the previous test is based on Royden's fourth edition "Real Analysis."

2-In addition, the general idea of the interval test $ (0.1) $ It is not clear to me, could anyone explain it, please?

3 interval for $ (- 1.0), $ Could someone help me complete it, please?