This shows you the differences between two versions of the page.
Both sides previous revision Previous revision | Next revision Both sides next revision | ||
math105-s22:notes:lecture_15 [2022/03/08 00:27] pzhou |
math105-s22:notes:lecture_15 [2022/03/08 00:28] pzhou |
||
---|---|---|---|
Line 27: | Line 27: | ||
If $f$ is not bounded, then we can split $f$ to be a bounded part and an unbounded part. For any $n>0$, we can define $f = f_n + g_n$, where $f_n = \max (f, n)$. Then, as $n \to \infty$, | If $f$ is not bounded, then we can split $f$ to be a bounded part and an unbounded part. For any $n>0$, we can define $f = f_n + g_n$, where $f_n = \max (f, n)$. Then, as $n \to \infty$, | ||
- | $$ \sum_i \int_{a_i}^b_i f(t) dt \leq \sum_i \int_{a_i}^b_i f_N(t) dt + \sum_i \int_{a_i}^b_i g_N(t) dt \leq N \delta + \int_a^b g_N \leq \epsilon/2 + \epsilon/2 = \epsilon $$ | + | $$ \sum_i \int_{a_i}^{b_i} f(t) dt \leq \sum_i \int_{a_i}^{b_i} f_N(t) dt + \sum_i \int_{a_i}^{b_i} g_N(t) dt \leq N \delta + \int_a^b g_N \leq \epsilon/2 + \epsilon/2 = \epsilon $$ |
Done for (a). | Done for (a). | ||