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I quote a theorem due to Lusin:

Let $X$ be a locally compact Hausdorff space and let $\mu$ be a regular Borel measure on $X$ such that $\mu(K)<\infty$ for every compact $K\subseteq X$. Suppose $f$ is a complex measurable function on $X$, $\mu(A)<\infty$, $f(x)=0$ if $x\in X\setminus A$, and $\epsilon>0$. Then there exists a continuous complex function $g$ on $X$ with compact support such that

 

$\mu(\{x:f(x)\neq g(x)\})<\epsilon$.

 

Furthermore, the function $g$ can be chosen such that

 

$sup_{x\in X}|g(x)|\leq sup_{x\in X}|f(x)|$.

As an immediate corollary (which is more relevant to your question), observe that:

If the hypotheses of Lusin's theorem are satisfied and if $|f|\leq 1$, then there is a sequence $\{g_n\}$ of continuous complex functions with compact support such that $|g_n|\leq 1$ for all $n$ and

 

$f(x)=\lim_{n \to \infty}g_n(x)$

 

almost everywhere with respect to $\mu$.

Note that the proof of Lusin's Theorem requires Urysohn's lemma for locally compact Hausdorff spaces, or at least a variation of it. (A very similar argument to that used to prove Urysohn's lemma for Normal Hausdorff spaces establishes the fact that any locally compact Hausdorff space is completely regular.) For more details on these results and their proofs, see Chapter 2 of the second edition of Walter Rudin's Real and Complex Analysis. (The results can be more precisely located on pages 56 and 57.)

I quote a theorem due to Lusin:

Let $X$ be a locally compact Hausdorff space and let $\mu$ be a regular Borel measure on $X$ such that $\mu(K)<\infty$ for every compact $K\subseteq X$. Suppose $f$ is a complex measurable function on $X$, $\mu(A)<\infty$, $f(x)=0$ if $x\in X\setminus A$, and $\epsilon>0$. Then there exists a continuous complex function $g$ on $X$ with compact support such that

$\mu(\{x:f(x)\neq g(x)\})<\epsilon$.

Furthermore, the function $g$ can be chosen such that

$sup_{x\in X}|g(x)|\leq sup_{x\in X}|f(x)|$.

As an immediate corollary (which is more relevant to your question), observe that:

If the hypotheses of Lusin's theorem are satisfied and if $|f|\leq 1$, then there is a sequence $\{g_n\}$ of continuous complex functions with compact support such that $|g_n|\leq 1$ for all $n$ and

$f(x)=\lim_{n \to \infty}g_n(x)$

almost everywhere with respect to $\mu$.

Note that the proof of Lusin's Theorem requires Urysohn's lemma for locally compact Hausdorff spaces, or at least a variation of it. (A very similar argument to that used to prove Urysohn's lemma for Normal Hausdorff spaces establishes the fact that any locally compact Hausdorff space is completely regular.) For more details on these results and their proofs, see Chapter 2 of the second edition of Walter Rudin's Real and Complex Analysis. (The results can be more precisely located on pages 56 and 57.)

I quote a theorem due to Lusin:

Let $X$ be a locally compact Hausdorff space and let $\mu$ be a regular Borel measure on $X$ such that $\mu(K)<\infty$ for every compact $K\subseteq X$. Suppose $f$ is a complex measurable function on $X$, $\mu(A)<\infty$, $f(x)=0$ if $x\in X\setminus A$, and $\epsilon>0$. Then there exists a continuous complex function $g$ on $X$ with compact support such that

$\mu(\{x:f(x)\neq g(x)\})<\epsilon$.

Furthermore, the function $g$ can be chosen such that

$sup_{x\in X}|g(x)|\leq sup_{x\in X}|f(x)|.

As an immediate corollary (which is more relevant to your question), observe that:

If the hypotheses of Lusin's theorem are satisfied and if $|f|\leq 1$, then there is a sequence $\{g_n\}$ of continuous complex functions with compact support such that $|g_n|\leq 1$ for all $n$ and

$f(x)=\lim_{n \to \infty}g_n(x)$

almost everywhere with respect to $\mu$.

Note that the proof of Lusin's Theorem requires Urysohn's lemma for locally compact Hausdorff spaces, or at least a variation of it. (A very similar argument to that used to prove Urysohn's lemma for Normal Hausdorff spaces establishes the fact that any locally compact Hausdorff space is completely regular.) For more details on these results and their proofs, see Chapter 2 of the second edition of Walter Rudin's Real and Complex Analysis. (The results can be more precisely located on pages 56 and 57.)

I quote a theorem due to Lusin:

Let $X$ be a locally compact Hausdorff space and let $\mu$ be a regular Borel measure on $X$ such that $\mu(K)<\infty$ for every compact $K\subseteq X$. Suppose $f$ is a complex measurable function on $X$, $\mu(A)<\infty$, $f(x)=0$ if $x\in X\setminus A$, and $\epsilon>0$. Then there exists a continuous complex function $g$ on $X$ with compact support such that

$\mu(\{x:f(x)\neq g(x)\})<\epsilon$.

Furthermore, the function $g$ can be chosen such that

$sup_{x\in X}|g(x)|\leq sup_{x\in X}|f(x)|$.

As an immediate corollary (which is more relevant to your question), observe that:

If the hypotheses of Lusin's theorem are satisfied and if $|f|\leq 1$, then there is a sequence $\{g_n\}$ of continuous complex functions with compact support such that $|g_n|\leq 1$ for all $n$ and

$f(x)=\lim_{n \to \infty}g_n(x)$

almost everywhere with respect to $\mu$.

Note that the proof of Lusin's Theorem requires Urysohn's lemma for locally compact Hausdorff spaces, or at least a variation of it. (A very similar argument to that used to prove Urysohn's lemma for Normal Hausdorff spaces establishes the fact that any locally compact Hausdorff space is completely regular.) For more details on these results and their proofs, see Chapter 2 of the second edition of Walter Rudin's Real and Complex Analysis. (The results can be more precisely located on pages 56 and 57.)

I quote a theorem due to Lusin:

Let $X$ be a locally compact Hausdorff space and let $\mu$ be a regular Borel measure on $X$ such that $\mu(K)<\infty$ for every compact $K\subseteq X$. Suppose $f$ is a complex measurable function on $X$, $\mu(A)<\infty$, $f(x)=0$ if $x\in X\setminus A$, and $\epsilon>0$. Then there exists a continuous complex function $g$ on $X$ with compact support such that

$\mu(\{x:f(x)\neq g(x)\})<\epsilon$.

Furthermore, the function $g$ can be chosen such that

$sup_{x\in X}|g(x)|\leq sup_{x\in X}|f(x)|.

As an immediate corollary (which is more relevant to your question), observe that:

If the hypotheses of Lusin's theorem are satisfied and if $|f|\leq 1$, then there is a sequence $\{g_n\}$ of continuous complex functions with compact support such that $|g_n|\leq 1$ for all $n$ and

$f(x)=\lim_{n \to \infty}g_n(x)$

almost everywhere with respect to $\mu$.

Note that the proof of Lusin's Theorem requires Urysohn's lemma. For more details on these results and their proofs, see Chapter 2 of the second edition of Walter Rudin's Real and Complex Analysis. (The results can be more precisely located on pages 56 and 57.)

I quote a theorem due to Lusin:

Let $X$ be a locally compact Hausdorff space and let $\mu$ be a regular Borel measure on $X$ such that $\mu(K)<\infty$ for every compact $K\subseteq X$. Suppose $f$ is a complex measurable function on $X$, $\mu(A)<\infty$, $f(x)=0$ if $x\in X\setminus A$, and $\epsilon>0$. Then there exists a continuous complex function $g$ on $X$ with compact support such that

$\mu(\{x:f(x)\neq g(x)\})<\epsilon$.

Furthermore, the function $g$ can be chosen such that

$sup_{x\in X}|g(x)|\leq sup_{x\in X}|f(x)|.

As an immediate corollary (which is more relevant to your question), observe that:

If the hypotheses of Lusin's theorem are satisfied and if $|f|\leq 1$, then there is a sequence $\{g_n\}$ of continuous complex functions with compact support such that $|g_n|\leq 1$ for all $n$ and

$f(x)=\lim_{n \to \infty}g_n(x)$

almost everywhere with respect to $\mu$.

Note that the proof of Lusin's Theorem requires Urysohn's lemma for locally compact Hausdorff spaces, or at least a variation of it. (A very similar argument to that used to prove Urysohn's lemma for Normal Hausdorff spaces establishes the fact that any locally compact Hausdorff space is completely regular.) For more details on these results and their proofs, see Chapter 2 of the second edition of Walter Rudin's Real and Complex Analysis. (The results can be more precisely located on pages 56 and 57.)