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I'm having trouble in establishing a few properties of lower and upper Lebesgue integral which are defined as $$\underline\int_{\mathbb{R}^d}f(x)dx=\sup_{h \leq f ~a.e.; ~h:simple}Simp \int_{\mathbb{R}^d}h(x)dx$$ $$\overline\int_{\mathbb{R}^d}f(x)dx=\inf_{h \geq f ~a.e.; ~h:simple}Simp \int_{\mathbb{R}^d}h(x)dx$$

The properties that I have to establish are as follows:

(I) Horizontal truncation: As $n \to \infty$, show that $\underline\int_{\mathbb{R}^d}min(f(x),n)dx$ converges to $\underline\int_{\mathbb{R}^d}f(x)dx$.

(II) Vertical truncation: As $n \to \infty$, show that $\underline\int_{\mathbb{R}^d}f(x)1_{\lvert x \rvert \leq n}dx$ converges to $\underline\int_{\mathbb{R}^d}f(x)dx$.

Do the horizontal and vertical truncation properties hold if the lower Lebesgue integral is replaced by upper Lebesgue integral?

(III) Reflection: If $f+g$ is simple function that is bounded with finite measure support (i.e. it is absolutely integrable), then $Simp \int_{\mathbb{R}^d}\{f(x)+g(x)\}dx=\underline \int_{\mathbb{R}^d}f(x)dx+\overline \int_{\mathbb{R}^d}g(x)dx$

My problems:

In (I), if $f$ is bounded then horizontal truncation is trivial. By monotonicity, it is easy to deduce that $\lim_{n \to \infty} \underline \int_{\mathbb{R}^d}min(f(x),n)dx \leq \underline\int_{\mathbb{R}^d}f(x)dx$ in all cases. But I cannot show the other inequality for the non-trivial case (i.e. when the function is not bounded).

In (II), I have to use the result $m(E \cap \{x:\lvert x \rvert \leq n\}) \to m(E)$ as $n \to \infty$ for any measurable set $E$. But I cannot connect Lebesgue measure to the definition of lower Lebesgue integral (and obviously to the definition of upper Lebesgue integral to check whether the property holds for upper Lebesgue integral).

In (III), I don't really have a clue. The lower and upper Lebesgue integral in the R.H.S. are giving me two opposite inequalities from which no conclusion can be drawn.

Any help would be greatly appreciated!

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Hint for (I): by definition of the lower integral, you can find a simple function $h \le f$ whose integral is close to the lower integral of $f$. Simple functions are bounded, hence for all sufficiently large $n$, you have $h \le \min(f,n)$ as well. So you should be able to conclude something about $\liminf_{n \to \infty} \underline{\int} \min(f, n)$.

Hint for (II): The connection is that Lebesgue measure appears in the definition of the simple integral. As before, choose a simple $h$ whose integral is close to that of $f$. How does the vertical truncation of $h$ compare to that of $f$? Now note that the vertical truncation of $h$ is also a simple function, and you should be able to say something about what happens to its simple integral when $n \to \infty$.

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  • $\begingroup$ Thank you for the answer. I get your points. Can you give me a hint for Part-III (Reflection property) as well? Also I've got no idea why they call it Reflection property, there must be some intuition behind it. $\endgroup$ – Roronoa Jan 4 '17 at 11:45
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For (3),

For $\forall 0\le f'\le f$, $f'$ is a simple function, since $f+g$ is simple , we observe that $g':=f+g-f'$ is a simple function such that $g'\ge g$. Obviously we have \begin{align} &\text {Simp} {\int_{\mathbb {R}^d}}f(x)+g(x)\ \mathrm {d}x=\text {Simp} {\int_{\mathbb {R}^d}}f'(x)+g'(x)\ \mathrm {d}x\\&=\text {Simp} {\int_{\mathbb {R}^d}}f'(x)\mathrm {d}x+\text {Simp} {\int_{\mathbb {R}^d}}g'(x)\mathrm {d}x\ge \text {Simp} {\int_{\mathbb {R}^d}}f'(x)\mathrm {d}x+\overline {\int_{\mathbb {R}^d}}g(x)\mathrm {d}x \end{align} Taking the supremum we soon have \begin{align} \text {Simp} {\int_{\mathbb {R}^d}}f(x)+g(x)\ \mathrm {d}x\ge \underline {\int_{\mathbb {R}^d}}f(x)\mathrm {d}x+\overline {\int_{\mathbb {R}^d}}g(x)\mathrm {d}x \end{align} The other direction can be obtained similarly. Thus we get the result.

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