Application of Monotone Convergence Theorem

Question

Suppose $f ∈ L^{1}([0, 1])$. Prove that $lim$ $ε→0^{+} \int_{[0,ε]} f dµ = 0$

My attempt at proof:

Let $B_N$ be an open ball of radius $N$ centred at origin.

$E_N:=$ {$x: f(x)\leq N$}

$f_N(x):=f(x) \chi_{E_N}$

$f_N$ is measurable. Also, $f_{N+1}\geq f_{N}$ and $lim _N\to \infty f_N(x)=f(x)$

By Monotone Convergence Theorem,

$lim _N\to \infty \int_{[0,1]} f_N d\mu=\ \int_{[0,1]} f d\mu$

$\implies 0\leq \int_{[0,1]} f(x)\ d\mu- \int_{[0,1]} f(x)\chi_{E_N}\ d\mu<\epsilon/2$ for some $\epsilon$

$\implies \int_{[0,1]} [f(x)-f_N(x)]\ d\mu<\epsilon/2$.

$\mu([0,\epsilon])=\epsilon$.

So, $\int_{[0,\epsilon]} f\ d\mu= \int_{[0,\epsilon]} [f-f_N]\ d\mu+\int_{[0,\epsilon]} f_N\ d\mu\leq \int_{[0,1]} [f-f_N]\ d\mu + \int_{[0,\epsilon]} f_N\ d\mu \leq \epsilon/2+N \epsilon$

Please suggest if the proof is correct.

Answer 1

Let us assume $f\ge 0$.

Consider the sequence $f_\epsilon = f\times 1_{(0,\epsilon)}$. $f_\epsilon$ decreases when $\epsilon\to 0$ as $f\ge 0$, and $f_\epsilon (x) \to 0$ pointwise.

Applying the monotonic convergence theorem, you get $$ \int_0^\epsilon fd\mu = \int f_\epsilon d\mu\to 0 $$

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When $f\in L^1$, write $f = f^+ - f^-$ with $f^\pm \ge 0$. And apply the first part twice.

Answer 2

This can be done without any of the convergence theorems. First, it's clear if $f$ is bounded. For the general $f\in L^1,$ let $t> 0.$ Then there is a simple function $s$ such that $\int_0^1|f-s|<t.$ Thus

$$\int_0^\epsilon|f| \le \int_0^\epsilon|f-s|+\int_0^\epsilon|s|< t + \int_0^\epsilon|s|.$$

This shows

$$\limsup_{\epsilon\to 0}\int_0^\epsilon|f| \le t.$$

Since $t>0$ is arbitrary, we have the desired result.