Let $\varepsilon >0$ exist a finite interval $[a,b]$ and a bounded function $h$ such that then $\int_{-\infty}^{\infty}{|f(x)-h(x)|}dx<\varepsilon$

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ Lebesgue integrable. Let $\varepsilon >0$ exist a finite interval $[a,b]$ and a bounded function $h$ such that $h(x)=0$ if $x\not\in[a,b]$ then

$$\int_{-\infty}^{\infty}{|f(x)-h(x)|}dx<\varepsilon.$$

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My attempt：

As $f$ is Lebesgue integrable then $f<\infty$.

Let $$g_n(x)=\begin{cases}f(x)&\text{if}\ x\in[a,b]\\0&\text{otherwise},\end{cases}$$

Note that $g_n\rightarrow f$, then apply the Dominate convergence theorem

$$\int_{a}^{b} g_n\,dx \to \int f\,dx$$

Here, I'm stuck. Can someone help me?

Answer 1

You have to find $[a,b]$ and $h$. Your argument does not work. Take $g_n(x)=fI_{\{x:|f(x)| \leq n\}} I_{[-n,n]}(x)$ and apply DCT. Take $h=g_n$, $a=-n$ and $b=n$ with $n$ sufficiently large.

Answer 2

$f$ is measurable and finite almost everywhere since $f$ is Lebesgue integrable. We can find a sequence of simple functions $\{f_n\}$ with compact supports, such that $f_n\to f$ almost everywhere and $|f_n|\le |f|$.

In order to satisfy the boundedness condition, we modify $f_n$ slightly:

Set $h_n=f_n\chi_{\{|f_n|\le n\}}$, then $h_n\to f$ a.e. and $|h_n|\le |f|$ but $h_n$ is bounded by $n$.

Since $f$ is integrable, dominated convergence theorem tells us that $$ \lim_{n\to\infty}\int_{\mathbb R}|f(x)-h_n(x)|dx=0 .$$

By the definition of limit, for every $\varepsilon>0$, we can find a $N>0$, such that for every $n\ge N$, $$\int_{\mathbb R}|f(x)-h(x)|dx<\varepsilon$$ where $h(x)\colon =h_N(x)$.

Note that $h(x)=h_N(x)$ has compact support, say $\{x\colon h(x)\ne 0\}\subset [a,b]$, then we have already found such a function $h$.