Handle finite integral of unbounded function I am trying to show that there exists a $\delta>0$ such that in the measure space $(X,\mathcal{A},\mu),\, u \in \mathcal{L}^1$:
$\forall E \in \mathcal{A}: \mu(E) < \delta \Rightarrow  |\int_E u\,d\mu|<  \frac{1}{100}$
I can show this if u is bounded. However, the problem is if u is unbounded. Then the integral can still be finite, e.g. $\frac{1}{\sqrt x}$. 
I can't find a delta when u is unbounded.
Any hint/help would be appreciated
 A: By considering $u^{+}$ and $u^{-}$ we can reduce the proof to the case when $u \geq 0$. 
Suppose there is no such $\delta$. Then for each $n$ there exists $E_n$ such that $\mu (E_n)<\frac 1 {2^{n}}$ and $\int_{E_n} u d\mu>\frac 1 {100}$.   Let $E=\lim\sup E_n$. Then $\mu(E)=0$, so  $\int_E u d\mu=0$. But $\int_E u d\mu = \lim \int_{\cup_{j\geq n} E_j} u  d\mu \geq \lim\int_{ E_n} u d \mu \geq \frac 1 {100}$ which is  contradiction. 
Proof of the fact that $\mu (E)=0$: $\mu (E)=\mu (\cap_n \cup_{j \geq n} E_j) =\lim_n \mu (\cup_{j \geq n} E_j)\leq \lim_n\sum_{j \geq n} \frac 1 {2^{n}} =0$. 
A: Since $u\in L^1$, $u$ is finite almost everywhere.
Therefore $|u|\land N:=min\{|u|, N\}$ monotonically increases and converges to $|u|$ a.e. as $N\rightarrow\infty$.
Choose sufficiently large $N$ so that $\int_{X}|u|d\mu-\int_{X}(|u|\land N)d\mu=\int_{X}(|u|-|u|\land N)d\mu<\epsilon$. Note that the integrand is always positive.
Since $\int_{E}(|u|\land N)d\mu\leq N\mu(E)$, you can now choose $E$ with sufficiently small $\mu(E)$ so that $\int_{E}(|u|\land N)d\mu<\epsilon$.
Combining the previous lines proves what you need. Just set $\epsilon=\frac{1}{200}$.
